Download App

Maths M.C.Q (1 Marks)

Straight Lines · CBSE STD 11 Science (CBSE - English Medium). 170 questions.

Questions

M.C.Q (1 Marks)

Take a timed test

170 questions · auto-graded multiple-choice test.

MCQ 11 Mark
Two lines are said to be perpendicular if the product of their slope is equal to:
  • ✓
    $-1$
  • B
    $0$
  • C
    $1$
  • D
    $\frac{1}{2}$
Answer
Correct option: A.
$-1$
When two lines are perpendicular, then the product of their slope is equal to $-1.$
If two lines are perpendicular with slope $m_1$ and $m_2$, then $m_1.m_2 = -1.$
View full question & answer→
MCQ 21 Mark
If slope of a line is $4$ and $y-$intercept made by the line is $2$ then the equation of line will be:
  • A
    $y = 4x - 2$
  • ✓
    $y = 4x + 2$
  • C
    $y = 2x + 4$
  • D
    $y = 2x - 4$
Answer
Correct option: B.
$y = 4x + 2$
Let general equation of line be $y = m \times x + c.$
Given, $m = 4$ and $c = 2.$
$\Rightarrow y = 4x + 2$
View full question & answer→
MCQ 31 Mark
Given the three straight lines with equations $5x + 4y = 0, x + 2y - 10 = 0$ and $2x + y + 5 = 0,$ then these lines are:
  • A
    None of these
  • B
    The sides of a right angled triangle
  • ✓
    Concurrent
  • D
    The sides of an equilateral triangle
Answer
Correct option: C.
Concurrent
View full question & answer→
MCQ 41 Mark
Choose the correct answer.
A point equidistant from the lines $4x + 3y + 10 = 0, 5x – 12y + 26 = 0$ and $7x + 24y – 50 = 0$ is:
  • A
    $(1, -1)$
  • B
    $(1, 1)$
  • ✓
    $(0, 0)$
  • D
    $(0, 1)$
Answer
Correct option: C.
$(0, 0)$
Given equation are
$4x + 3y + 10 = 0 .....(i)$
$5x - 12y + 26 = 0 .....(ii)$
and $7x + 27y - 50 = 0 .....(iii)$
Let $(x_1, y_1)$ be any point equidistant from eq. $(i),$ eq. $(ii)$ and eq. $(iii).$
Distance of $(x_1, y_1)$ from eq. $(i)$
$=\Big|\frac{4\text{x}_1+3\text{y}_1+10}{\sqrt{16+9}}\Big|=\Big|\frac{4\text{x}_1+3\text{y}_1+10}{5}\Big|$
Distance of $(x_1, y_1)$ from eq. $(iii)$
$=\Big|\frac{5\text{x}_1-12\text{y}_1+26}{\sqrt{25+144}}\Big|=\Big|\frac{5\text{x}_1-12\text{y}_1+26}{13}\Big|$
Distance of $(x_1, y_1)$ from eq. $(iii)$
$=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{\sqrt{49+576}}\Big|=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{25}\Big|$
If the point $(x_1, y_1)$ is equidistant from the given lines, then
$\Big|\frac{4\text{x}_1+3\text{y}_1+10}{5}\Big|=\Big|\frac{5\text{x}_1-12\text{y}_!+26}{13}\Big|=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{25}\Big|$
We see that putting $x_1 = 0$ and $y_1= 0$, the above relation is satisfied i.e.,
$=\frac{10}{5}=\frac{26}{13}=\frac{50}{25}=2$
Hence, the correct option is $(c).$
View full question & answer→
MCQ 51 Mark
Find the equation of line parallel to $4x + y = 2$ and pass through $(2, 5):$
  • ✓
    $4x + y - 13 = 0$
  • B
    $4x + y + 13 = 0$
  • C
    $4x - y - 13 = 0$
  • D
    $4x - y + 13 = 0$
Answer
Correct option: A.
$4x + y - 13 = 0$
Line $4x + y = 2$ has
slope $-4$ Line parallel to it has
slope $-4$ and pass through $(2, 5)$
so equation will be $y - 5 = (-4) (x - 2)$
$\Rightarrow 4x + y - 13 = 0$
View full question & answer→
MCQ 61 Mark
If slope of a line is positive then its inclination is:
  • A
    Right angle.
  • ✓
    Acute angle.
  • C
    Obtuse angle.
  • D
    Zero.
Answer
Correct option: B.
Acute angle.
If inclination is $\alpha$ slope is given by $\tan\alpha$ Given that slope of line is positive which means $\tan\alpha$ is positive.
We know, $\tan\alpha$ is positive in $1^{st}$ quadrant
i.e. $\alpha$ should be acute angle.
View full question & answer→
MCQ 71 Mark
What is the distance of $(5, 12)$ from origin?
  • A
    $6$ units.
  • B
    $8$ units.
  • C
    $10$ units.
  • ✓
    $13$ units.
Answer
Correct option: D.
$13$ units.
We know, distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{({\text{x}}_{1}-{\text{x}}_{2})^{2}+({\text{y}}_{1}-{\text{y}}_{2})^{2}}$
So, distance between $(5, 12)$ from origin $(0, 0)$ is
$\sqrt{({5-0})^{2}+({12-0})^{2}}$
$= \sqrt{({5})^{2}+({12})^{2}}$
$=13\text{ unit}.$
View full question & answer→
MCQ 81 Mark
Three vertices of a parallelogram taken in order are $(-1, -6), (2, -5)$ and $(7, 2).$ The fourth vertex is:
  • A
    $(1, 4)$
  • ✓
    $(4, 1)$
  • C
    $(1, 1)$
  • D
    $(4, 4)$
Answer
Correct option: B.
$(4, 1)$
Let $A(-1, -6), B(2, -5)$ and $C(7, 2)$ be the given vertex.
Let $D(h, k)$ be the fourth vertex.
The midpoints of $AC$ and $BD$ are $(3, -2)$ and $\Big(\frac{2+\text{h}}{2},\frac{-5+\text{k}}{2}\Big)$ respectively.
We know that the diagonals of a parallelogram bisect each other.
$\therefore3=\frac{2+\text{h}}{2}$ and $-2=\frac{-5+\text{k}}{2}$
$\Rightarrow\text{h}=4$ and $\text{k}=1$
View full question & answer→
MCQ 91 Mark
If slope of a line is $\frac{2}{3}$ then find the slope of line perpendicular to it:
  • ✓
    $\frac{-3}{2}$
  • B
    $\frac{3}{2}$
  • C
    $\frac{2}{3}$
  • D
    $\frac{-2}{3}$
Answer
Correct option: A.
$\frac{-3}{2}$
If lines with slopes $m_1$ and $m_2$ are perpendicular then $m_1 \times m_2 = -1$.
If $\text{m}_{1} = \frac{2}{3}$ then $\text{m}_{2} = \frac{-3}{2} $
View full question & answer→
MCQ 101 Mark
The point which divides the join of $(1, 2)$ and $(3, 4)$ externally in the ratio $1 : 1:$
  • A
    Lies in the $III$ quadrant.
  • B
    Lies in the $II$ quadrant.
  • C
    Lies in the $I$ quadrant.
  • ✓
    Cannot be found.
Answer
Correct option: D.
Cannot be found.
The point which divides the join of $(1, 2)$ and $(3, 4)$ externally in the ratio $1 : 1$ is
$\Big(\frac{1\times3-1\times1}{1-1},\frac{1\times4-1\times2}{1-1}\Big)$
which is not defined.
Therefore, it is not possible to externally divide the line joining two points in the ratio $1 : 1$
View full question & answer→
MCQ 111 Mark
Find the equation of line parallel to $y-$axis and passing through $(3, 4):$
  • ✓
    $x = 3$
  • B
    $x = 4$
  • C
    $y = 4$
  • D
    $y = 3$
Answer
Correct option: A.
$x = 3$
Let general equation of line be $y = m (x - d)$
$\Rightarrow\text{x} = \frac{\text{y}}{\text{m + d}}$
Since line is parallel to $y-$axis
so, $\text{m}=\frac{1}{0}$ or $\frac{1}{\text{m}} =0$
$\Rightarrow x = d$
$\Rightarrow x = 3$ by substituting the point $(3, 4).$
View full question & answer→
MCQ 121 Mark
Choose the correct answer. Equation of the line passing through $(1, 2)$ and parallel to the line $y = 3x - 1$ is:
  • A
    $y + 2 = x + 1$
  • B
    $y + 2 = 3 (x + 1)$
  • ✓
    $y - 2 = 3 (x - 1)$
  • D
    $y - 2 = x - 1$
Answer
Correct option: C.
$y - 2 = 3 (x - 1)$
Given equation is $y = 3x - 1$
Slope $= 3$
Slope of the line passing through the given point $(1, 2)$ and parallel to the given line $= 3$
So, the equation of the required line is
$y - 2 = 3(x - 1)$
Hence, the correct option is $(c).$
View full question & answer→
MCQ 131 Mark
Choose the correct answer. The tangent of angle between the lines whose intercepts on the axes are $a, -b$ and $b, -a,$ respectively, is
  • A
    $\frac{\text{a}^2-\text{b}^2}{\text{ab}}$
  • B
    $\frac{\text{b}^2-\text{a}^2}{2}$
  • ✓
    $\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$
  • D
    None of these.
Answer
Correct option: C.
$\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$
Intercepts of line are $a$ and $-b;$
i.e., line passes through the points $(a, 0), (0, -b).$
$\therefore$ Slope of line, $\text{m}_1=\frac{-\text{b}-0}{0-\text{a}}=\frac{\text{b}}{\text{a}}$
Intercepts of line are $b, -a;$
i.e., line passes through the points $(b, 0), (0, -a).$
$\therefore$ Slope of line, $\text{m}_2=\frac{-\text{a}-0}{0-\text{b}}=\frac{\text{a}}{\text{b}}$
If $\theta$ is the angle between the lines, then
$\tan=\theta=\frac{\frac{\text{b}}{\text{a}}-\frac{\text{a}}{\text{b}}}{1+\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{a}}}=\frac{\frac{\text{b}^2-\text{a}^2}{\text{ab}}}{2}=\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$
View full question & answer→
MCQ 141 Mark
Find slope of line passing through origin and $(3, 6):$
  • ✓
    $2$
  • B
    $​​3$
  • C
    $\frac{1}{3}$
  • D
    $\frac{1}{2}$
Answer
Correct option: A.
$2$
We know, slope of line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{(\text{y}_{2}-\text{y}_{1})}{(\text{x}_{2}-\text{x}_{1})}$
So, slope of line joining $(0, 0)$ and $(3, 6)$ is $\frac{(6-0)}{(3-0)} = \frac{6}{3} = 2$
View full question & answer→
MCQ 151 Mark
A triangle $\text{ABC}$ is right angled at $A$ has points $A$ and $B$ as $(2, 3)$ and $(0, -1)$ respectively. If $BC = 5,$ then point $C$ may be:
  • A
    $(-4, 2)$
  • B
    $(4, -2)$
  • ✓
    $(0, 4)$
  • D
    $(0, -4)$
Answer
Correct option: C.
$(0, 4)$
View full question & answer→
MCQ 161 Mark
$L$ is a variable line such that the algebraic sum of the distances of the points $(1, 1), (2, 0)$ and $(0, 2)$ from the line is equal to zero. The line $L$ will always pass through:
  • ✓
    $(1, 1)$
  • B
    $(2, 1)$
  • C
    $(1, 2)$
  • D
    none of these.
Answer
Correct option: A.
$(1, 1)$
Let $ax + by + c = 0$ be the variable line.
It is given that the algebraic sum of the distances of the points $(1, 1), (2, 0)$ and $(0, 2)$ from the line is equal to zero
$\therefore \ \frac{\text{a}+\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{2\text{a}+0+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{0+2\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}=0$
$\Rightarrow3\text{a}+3\text{b}+3\text{c}=0$
$\Rightarrow\text{a}+\text{b}+\text{c}=0$
Substituting $c = -a - b$ in $ax + by + c = 0,$ we get:
$\text{ax}+\text{by}-\text{a}-\text{b}=0$
$\Rightarrow\text{a}(\text{x}-1)+\text{b}(\text{y}-1)=0$
$\Rightarrow(\text{x}-1)+\frac{\text{a}}{\text{b}}(\text{y}-1)=0$
This line is of the form $\text{L}_1+\lambda\text{L}_2=0$
which passes through the intersection of $L_1 =0$ and $L_2=0$,
i.e. $x - 1 = 0$ and $y - 1 = 0.$
$\Rightarrow x = 1, y = 1$
View full question & answer→
MCQ 171 Mark
Equation of horizontal line below $x-$axis at $5$ units from $x-$axis is:
  • A
    $x = 5$
  • B
    $x = -5$
  • C
    $y = 5$
  • ✓
    $y = -5$
Answer
Correct option: D.
$y = -5$
Equation of $x-$axis is $y = 0.$ Horizontal line is parallel to $x-$axis and below it by $5$ units
so, equation of line is $y = -5.$
View full question & answer→
MCQ 181 Mark
Choose the correct answer. The equation of the line passing through the point $(1, 2)$ and perpendicular to the line $x + y + 1 = 0$ is:
  • A
    $y - x + 1 = 0$
  • ✓
    $y - x - 1 = 0$
  • C
    $y - x + 2 = 0$
  • D
    $y - x - 2 = 0$
Answer
Correct option: B.
$y - x - 1 = 0$
Slope of the given line $+1 = 0$ is $-1.$
So, slope of line perpendicular to above line is $1.$
Line passes through the point $(1, 2).$
Therefore, equation of the required linens:
$\Rightarrow y - 2 = 1(x - 1)$
$\Rightarrow y - x - 1 = 0.$
View full question & answer→
MCQ 191 Mark
Choose the correct answer. The equations of the lines passing through the point $(1, 0)$ and at a distance $\frac{\sqrt{3}}{2}$ from the origin, are
  • ✓
    $\sqrt{3}{x}+{y}-\sqrt{3}=0,\sqrt{3}{x}-{y}-\sqrt{3}=0$
  • B
    $\sqrt{3}{x}+{y}+\sqrt{3}=0,\sqrt{3}{x}-{y}+\sqrt{3}=0$
  • C
    ${x}+\sqrt{3}{y}-\sqrt{3}=0,{x}-\sqrt{3}{y}-\sqrt{3}=0$
  • D
    None of these.
Answer
Correct option: A.
$\sqrt{3}{x}+{y}-\sqrt{3}=0,\sqrt{3}{x}-{y}-\sqrt{3}=0$
Equation of any line passing through $(1, 0)$ is
$\Rightarrow y - 0 = m(x - 1)$
$\Rightarrow mx - y - m = 0$
Distance of the line from origin is $\frac{\sqrt{3}}{2}$
$\therefore \frac{\sqrt{3}}{2} =\Big|\frac{\text{m}\times0-0-\text{m}}{\sqrt{1+\text{m}^2}}\Big|$
$\Rightarrow \frac{\sqrt{3}}{2}=\Big|\frac{-\text{m}}{\sqrt{1+\text{m}^2}}\Big|$
Squaring both sides, we get
$\frac{3}{4}=\frac{\text{m}^2}{1+\text{m}^2}$
$\Rightarrow 4\text{m}^2=3+3\text{m}^2$
$\Rightarrow 4\text{m}^2-3\text{m}^2=3$
$\Rightarrow \text{m}^2=3$
$\therefore \text{m}=\pm\sqrt{3}$
$\therefore$ Required equations are
$\pm\sqrt{3}\text{x}-\text{y}\mp\sqrt{3}=0$
i.e., $\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0=0$ and $-\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0=0$
$\Rightarrow \sqrt{3}\text{x}+\text{y}-\sqrt{3}=0$
Hence, the correct option is $(a).$
View full question & answer→
MCQ 201 Mark
If the points $A (1, 2), B (2, 4)$ and $C (3, a)$ are collinear, what is the length $BC?$
  • A
    $2$ unit
  • B
    $3$ unit
  • ✓
    $5$ unit
  • D
    $5$ unit
Answer
Correct option: C.
$5$ unit
View full question & answer→
MCQ 211 Mark
If the two lines with slope $m_1$ and $m_2$ are perpendicular then their slopes has relation:
  • A
    $ m_1+m_2=1 $
  • B
    $ m_1 \times m_2=1 $
  • ✓
    $ m_1 \times m_2=-1 $
  • D
    $ m_1+m_2=-1 $
Answer
Correct option: C.
$ m_1 \times m_2=-1 $
If the two lines are perpendicular then if one line form angle $\alpha$ with positive $x-$axis then the other line form angle $90^\circ + \alpha$
If $\text{m}_{1} = \tan \alpha$ then $m_2$ will be $\tan (90^\circ + \alpha) = -\cot\alpha = \frac{-1}{\tan\alpha}$
$\Rightarrow m_1 \times m_2=-1 $
View full question & answer→
MCQ 221 Mark
If $P (1, 2), Q (3, 5), R (7, 9)$ form a triangle then find the equation of median through $P:$
  • ✓
    $5x - 4y + 3 = 0$
  • B
    $5x + 4y + 3 = 0$
  • C
    $5x - 4y - 3 = 0$
  • D
    $5x + 4y - 3 = 0$
Answer
Correct option: A.
$5x - 4y + 3 = 0$
Midpoint of $QR$ line is $\Big(\frac{3+7}{2},\frac{5+9}{2}\Big) = (5, 7)$
Equation of line joining $(1, 2)$ and $(5, 7)$ is $\frac{\text{y - 2}}{7 - 2} = \frac{\text{x - 1}}{5-1}$
$\Rightarrow\frac{\text{y}-2}{5} = \frac{\text{x}-1}{4}$
$\Rightarrow 4y - 8 = 5x - 5$
$\Rightarrow 5x - 4y + 3 = 0.$
View full question & answer→
MCQ 231 Mark
A line passes through $(2, 2)$ and is perpendicular to the line $3x + y = 3.$ Its $y-$intercept is:
  • A
    $\frac{1}{3}$
  • B
    $\frac{2}{3}$
  • C
    $1$
  • ✓
    $\frac{4}{3}$
Answer
Correct option: D.
$\frac{4}{3}$
View full question & answer→
MCQ 241 Mark
The acute angle between the medians drawn from the acute angles of a right angled isosceles triangle is:
  • A
    $\cos^{-1}\big(\frac{2}{3}\big)$
  • B
    $\cos^{-1}\big(\frac{3}{4}\big)$
  • ✓
    $\cos^{-1}\big(\frac{4}{5}\big)$
  • D
    $\cos^{-1}\big(\frac{5}{6}\big)$
Answer
Correct option: C.
$\cos^{-1}\big(\frac{4}{5}\big)$
Let the coordinates of the right$-$angled isosceles triangle be $O(0, 0), A(a, 0)$ and $B(0, a).$

Here, $BD$ and $AE$ are the medians drawn from the acute angles $B$ and $A$, respectively.
$\therefore$ Slope of $BD = m_1$
$=\frac{0-\text{a}}{\frac{\text{a}}{2}-0}$
$=-\frac{1}{2}$
Let $\theta$ be the angle between $BD$ and $AE.$
$\tan\theta=\Big|\frac{-2+\frac{1}{2}}{1+1}\Big|$
$=\frac{3}{4}$
$\Rightarrow\cos\theta=\frac{4}{\sqrt{3^2+4^2}}$
$\Rightarrow\cos\theta=\frac{4}{5}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{4}{5}\Big)$
Hence, the acute angle between the medians is $\cos^{-1}\Big(\frac{4}{5}\Big).$
View full question & answer→
MCQ 251 Mark
$.....$ is the midpoint of $(1, 2)$ and $(5, 8):$
  • A
    $(2, 5)$
  • ✓
    $(3, 5)$
  • C
    $(5, 2)$
  • D
    $(5, 3)$
Answer
Correct option: B.
$(3, 5)$
We know, midpoint of $(x_1, y_1)$ and $(x_2, y_2)$ is $\Big(\frac{{\text{x}}_{1}+{\text{x}}_{2}}{2}, \frac{{\text{y}}_{1}+{\text{y}}_{2}}{2}\Big)$
So, midpoint of $(1, 2)$ and $(5, 8)$ is $\Big(\frac{1+5}{2}, \frac{2+8}{2}\Big)$ is $(3, 5)$
View full question & answer→
MCQ 261 Mark
If line joining $(1, 2)$ and $(5, 7)$ is parallel to line joining $(3, 4)$ and $(11, x):$
  • A
    $10$
  • B
    $11$
  • C
    $12$
  • ✓
    $14$
Answer
Correct option: D.
$14$
We know, slope of line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{(\text{y}_{2}-\text{y}_{1})}{(\text{x}_{2}-\text{x}_{1})}$
Lines are parallel means slope is equal.
$\Rightarrow\frac{(\text{x}-4)}{(11-3)} =\frac {(7-2)}{(5-1)}$
$\text{x}-4 = 5\times\frac{8}{4} = 10$
$\Rightarrow x = 14$
View full question & answer→
MCQ 271 Mark
If $(-4, 5)$ is one vertex and $7x - y + 8 = 0$ is onediagonal of a square, then the equation of second diagonal is:
  • A
    $x + 3y = 21$
  • B
    $2x - 3y = 7$
  • ✓
    $x + 7y = 31$
  • D
    $2x + 3y = 21$
Answer
Correct option: C.
$x + 7y = 31$
View full question & answer→
MCQ 281 Mark
Slope of a line is given by if inclination of line is $\alpha$:
  • A
    $\sin\alpha$
  • B
    $\cos\alpha$
  • ✓
    $\tan\alpha$
  • D
    $\cot\alpha$
Answer
Correct option: C.
$\tan\alpha$
Slope of a line is given by $\tan\alpha$ if inclination of line is $\alpha$. Slope is denoted by tangent of the inclination angle.
View full question & answer→
MCQ 291 Mark
Equation of vertical line to the right of $y-$axis at $5$ units from $y-$axis is:
  • ✓
    $x = 5$
  • B
    $x = -5$
  • C
    $y = 5$
  • D
    $y = -5$
Answer
Correct option: A.
$x = 5$
Equation of $y-$axis is $x = 0.$ Vertical line is parallel to $y-$axis and to the right by $5$ units.
so, equation of line is $x = 5.$
View full question & answer→
MCQ 301 Mark
In what ratio does the line $y - x + 2 = 0$ cut the line joining $(3, -1)$ and $( 8, 9)?$
  • ✓
    $2 : 3$
  • B
    $3 : 2$
  • C
    $3 : -2$
  • D
    $1 : 2$
Answer
Correct option: A.
$2 : 3$
View full question & answer→
MCQ 311 Mark
The medians $AD$ and $BE$ of a triangle with vertices $A(0, b), B(0, 0)$ and $C(a, 0)$ are perpendicular to each other, if
  • A
    $\text{a}=\frac{\text{b}}{2}$
  • B
    $\text{b}=\frac{\text{a}}{2}$
  • C
    $\text{ab}=1$
  • ✓
    $\text{a}=\pm\sqrt{2}\text{b}$
Answer
Correct option: D.
$\text{a}=\pm\sqrt{2}\text{b}$
The midpoints of $BC$ and $AC$ are $\text{D}\Big(\frac{\text{a}}{2},0\Big)$ and $\text{E}\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$
Slope of $\text{AD}=\frac{0-\text{b}}{\frac{\text{a}}{2}-0}$
Slope of $\text{BE}=\frac{-\frac{\text{b}}{2}}{\frac{\text{-a}}{2}}$
It is given that the medians are perpendicular to each other.
$\frac{0-\text{b}}{\frac{\text{a}}{2}-0}\times\frac{-\frac{\text{b}}{2}}{-\frac{\text{a}}{2}}=-1$
$\Rightarrow\text{a}=\pm\sqrt{2}\text{b}$
View full question & answer→
MCQ 321 Mark
Choose the correct answer. Slope of a line which cuts off intercepts of equal lengths on the axes is:
  • ✓
    $-1$
  • B
    $-0$
  • C
    $2$
  • D
    $\sqrt{3}$
Answer
Correct option: A.
$-1$
Intercept form of a line is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
$\Rightarrow \frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{a}}=1(\because \text{a}=\text{b})$
$\Rightarrow x + y = a$
$\Rightarrow y =- -x + a$
$\therefore$ Slope is $-1$
Hence, the correct option is $(a).$
View full question & answer→
MCQ 331 Mark
If the two lines are perpendicular then difference of their inclination angle is:
  • A
    $45^\circ$
  • B
    $60^\circ$
  • ✓
    $90^\circ$
  • D
    $180^\circ$
Answer
Correct option: C.
$90^\circ$
If the two lines are perpendicular then if one line form angle $\alpha$ with positive $x-$axis then the other line form angle $90^\circ+\alpha$
View full question & answer→
MCQ 341 Mark
If the area of the triangle with vertices $(x, 0), (1, 1)$ and $(0, 2)$ is $4$ square unit, then the value of $x$ is:
  • A
    $-2$
  • B
    $-4$
  • ✓
    $-6$
  • D
    $8$
Answer
Correct option: C.
$-6$
View full question & answer→
MCQ 351 Mark
What is the inclination of a line which is parallel to $x-$axis?
  • ✓
    $0^\circ$
  • B
    $180^\circ$
  • C
    $45^\circ$
  • D
    $90^\circ$
Answer
Correct option: A.
$0^\circ$
If a line is parallel to $x-$axis then angle formed by it with $x-$axis is zero.
So, its inclination is zero.
View full question & answer→
MCQ 361 Mark
The line segment joining the points $(-3, -4)$ and $(1, -2)$ is divided by $y-$axis in the ratio:
  • A
    $1 : 3$
  • B
    $2 : 3$
  • ✓
    $3 : 1$
  • D
    $3 : 2$
Answer
Correct option: C.
$3 : 1$
Let the points $(-3, -4)$ and $(1, -2)$ be divided by $y-$axis at $(0, t)$ in the ratio $m : n.$
$\therefore\Big(\frac{\text{m}-3\text{n}}{\text{m}+\text{n}},\frac{-2\text{m}-4\text{n}}{\text{m}+\text{n}}\Big)=(0,\text{t})$
$\Rightarrow0=\frac{\text{m}-3\text{n}}{\text{m}+\text{n}}$
$\Rightarrow\text{m}:\text{n}=3:1$
View full question & answer→
MCQ 371 Mark
If a line with slope $m$ makes $x-$intercept $d$. Then equation of the line is:
  • A
    $y = m(d - x)$
  • ✓
    $y = m(x - d)$
  • C
    $y = m(x + d)$
  • D
    $y = mx + d$
Answer
Correct option: B.
$y = m(x - d)$
View full question & answer→
MCQ 381 Mark
The condition for the points $(x, y), (-2, 2)$ and $(3, 1)$ to be collinear is:
  • ✓
    $x + 5y = 8$
  • B
    $x + 5y = 6$
  • C
    $5x + y = 8$
  • D
    $5x + y = 6$
Answer
Correct option: A.
$x + 5y = 8$
$x (2 - 1) -1(1 - y) + 3 (y - 2) = 0$ or $x + 5y = 8$
View full question & answer→
MCQ 391 Mark
If equation of a line is $y = 3x - 4$ then find the slope of line:
  • ✓
    $3$
  • B
    $-3$
  • C
    $4$
  • D
    $-4$
Answer
Correct option: A.
$3$
Comparing the above equation with general equation $y = m × x + c,$
$m = 3$ which is the slope of line.
View full question & answer→
MCQ 401 Mark
The equation of a line that passes through the points $(1, 5)$ and $(2, 3)$ is:
  • ✓
    $2x + y - 7 = 0$
  • B
    $2x - y - 7 = 0$
  • C
    $x + 2y - 7 = 0$
  • D
    $2x + y + 7 = 0$
Answer
Correct option: A.
$2x + y - 7 = 0$
We know that the equation of a line passes through two points $(x_1, y_1)$ and $(x_2, y_2)$ is
$\frac{(\text{y}-\text{y}_{1})}{(\text{x}\text{-x}_{1})} = \frac{(\text{y}_{2}-\text{y}_{1})}{(\text{x}_{2}-\text{x}_{1})}$
$(x_1, y_1) = (1, 5)$
$(x_2, y_2) = (2, 3)$
Now, substitute the values in the formula, we get
$\frac{(\text{y}-5)}{(\text{x}-1)} = \frac{(3-5)}{(2-1)}$
$\frac{(\text{y}-5)}{(\text{x}-1)} = \frac{(-2)}{(1)}$
$y - 5 = -2(x - 1)$
$y - 5 = -2x + 2$
$2x + y - 5 - 2 = 0$
$2x + y - 7 = 0$
$\therefore$ The equation of a line that passes through the points $(1, 5)$ and $(2, 3)$ is $2x + y - 7 = 0.$
View full question & answer→
MCQ 411 Mark
The locus of a point, whose abscissa and ordinate are always equal is:
  • ✓
    $x - y = 0$
  • B
    $x + y = 1$
  • C
    $x + y + 1 = 0$
  • D
    None of the above
Answer
Correct option: A.
$x - y = 0$
Let the abscissa and ordinate of a point $“P”$ be $(x, y)$
Given condition: Abscissa $=$ Ordinate
$\text{(i.e) x = y}$
The locus of a point is $x - y = 0.$
View full question & answer→
MCQ 421 Mark
Find the equation of line parallel to $4x + y = 2$ and pass through $(2, 5):$
  • ✓
    $4x + y - 13 = 0$
  • B
    $4x + y + 13 = 0$
  • C
    $4x - y - 13 = 0$
  • D
    $4x - y + 13 = 0$
Answer
Correct option: A.
$4x + y - 13 = 0$
Line $4x + y = 2$ has slope $-4.$ Line parallel to it has slope $-4$ and pass through $(2, 5)$
so equation will be $y - 5 = (-4) (x - 2)$
$\Rightarrow 4x + y - 13 = 0$
View full question & answer→
MCQ 431 Mark
Choose the correct answer.
A line cutting off intercept $-3$ from the $y-$axis and the tengent at angle to the xaxis is $\frac{3}{5}$, its equation is:
  • ✓
    $5y - 3x + 15 = 0$
  • B
    $3y - 5x + 15 = 0$
  • C
    $5y - 3x - 15 = 0$
  • D
    None of these.
Answer
Correct option: A.
$5y - 3x + 15 = 0$
Since the lines cut off intercepts $-3$ on $y-$axis then the line is passing through the point $(0, -3).$
Given that: $\tan\theta=\frac{3}{5}$
$\Rightarrow$ Slope of the line $\text{m}=\frac{3}{5}$
So, the equation of the line is
$y - y_1 = m(x - x_1)$
$\Rightarrow \text{y}+ 3 = \frac{3}{5}(\text{x} -0)$
$\Rightarrow 5y + 15 = 3x$
$\Rightarrow 3x - 5y - 15 = 0$
$\Rightarrow 5y - 3x + 15 = 0$
Hence, the correct option is $(a).$
View full question & answer→
MCQ 441 Mark
The number of real values of $\lambda$ for which the lines $\text{x} - 2\text{y} + 3 = 0, \lambda\text{x} + 3\text{y} + 1 = 0$ and $4\text{x} - \lambda\text{y} + 2 = 0$ are concurrent is:
  • ✓
    $0$
  • B
    $1$
  • C
    $2$
  • D
    Infinite.
Answer
Correct option: A.
$0$
$\text{x} - 2\text{y} + 3 = 0 \ ...(\text{i})$
$\lambda\text{x} + 3\text{y} + 1 = 0 \ ...(\text{ii})$
$4\text{x} - \lambda\text{y} + 2 = 0 \ ...(\text{iii})$
It is given that $(1), (2)$ and $(3)$ are concurrent.
$\therefore\begin{vmatrix} 1&-2&3\\\lambda&3&1\\4&-\lambda&2\end{vmatrix}=0$
$\Rightarrow(6+\lambda)+2(2\lambda-4)+3(-\lambda^2-12)=0$
$\Rightarrow6+\lambda+4\lambda-8-3\lambda^2-36=0$
$\Rightarrow5\lambda-3\lambda^2-38=0$
$\Rightarrow3\lambda^2-5\lambda+38=0$
The discriminant of this equation is $25 - 4 \times 3 \times 38 = -431$
Hence, there is no real value of $\lambda$ for which the lines $\text{x} - 2\text{y} + 3 = 0, \lambda\text{x} + 3\text{y} + 1 = 0$ and $4\text{x} - \lambda\text{y} + 2 = 0$ are concurrent.
View full question & answer→
MCQ 451 Mark
Find slope of line if inclination made by the line is 60°.
  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{\sqrt{3}}$
  • ✓
    $\sqrt{3}$
  • D
    $1$
Answer
Correct option: C.
$\sqrt{3}$
Slope of a line is given by $\tan\alpha$ if inclination of line is $\alpha$ If inclination is $60^\circ$ the slope is $\tan 60^\circ = \sqrt{3}$
View full question & answer→
MCQ 461 Mark
If the point $(5, 2)$ bisects the intercept of a line between the axes, then its equation is:
  • A
    $5x + 2y = 20$
  • ✓
    $2x + 5y = 20$
  • C
    $5x - 2y = 20$
  • D
    $2x - 5y = 20$
Answer
Correct option: B.
$2x + 5y = 20$
Let the equation of the line be $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
The coordinates of the intersection of this line with the coordinate axes are $(a, 0)$ and $(0, b).$
The midpoint of $(a, 0)$ and $(0, b)$ is $\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$
According to the question:
$\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)=(5,2)$
$\Rightarrow\frac{\text{a}}{2}=5,\frac{\text{b}}{2}=2$
$\Rightarrow\text{a}=10,\text{b}=4$
The equation of the required line is given below:
$\frac{\text{x}}{10}+\frac{\text{y}}{4}=1$
$\Rightarrow2\text{x}+5\text{y}=20$
View full question & answer→
MCQ 471 Mark
Choose the correct answer.
One vertex of the equilateral triangle with centroid at the origin and one side as $x + y - 2 = 0$ is:
  • A
    $(-1, -1)$
  • B
    $(2, 2)$
  • ✓
    $(-2, -2)$
  • D
    $(2, -2)$
Answer
Correct option: C.
$(-2, -2)$
Let $\text{ABC}$ be the equilateral triangle with vertex $A(h, k).$
Also, centroid is $G(0, 0).$

Now, $\text{AG}\bot\text{BC}$
Slope of line $BC$ or $x + y - 2 = 0$ is $-1.$
$\therefore$ Slope of $\text{AG},\frac{\text{k}}{\text{h}}=1$ or $h = k.$
Now distance of origin from $\text{BC}=\frac{|0+0-2}{\sqrt{1^2+1^2}|}=\sqrt{2}$
$\therefore$ Distance of $A$ form $\text{BC}=3\sqrt{2}=\frac{|\text{h}+\text{k}-2|}{\sqrt{1^2+1^2}}$
$\therefore |h + k - 2| = 6$
$\Rightarrow h + k - 8 = 0$ or $h + k + 4 = 0$
$\Rightarrow h + k - 8 = 0$ or $h + k + 4 = 0$
$\Rightarrow h = 4$ or $h = -2$
$\therefore$ Vertex is $(-2, -2).$
View full question & answer→
MCQ 481 Mark
What is the distance between $(1, 3)$ and $(5, 6):$
  • A
    $3$ units
  • B
    $4$ units
  • ✓
    $5$ units
  • D
    $25$ units
Answer
Correct option: C.
$5$ units
We know, distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(\text{x}_{1}−\text{x}_{2})^2+(\text{y}_{1}−\text{y}_{2})^2.}$
So, distance between $(1, 3)$ and $(5, 6)$ is $\sqrt{{(1-5)}^2+{(3-6)}^2}$
$= (4)^2+ (3)^2$
$= 5$ units
View full question & answer→
MCQ 491 Mark
Choose the correct answer.
If the line $\frac{\text{x}}{\text{a}} + \frac{\text{y}}{\text{b}} =1$ passes through the points $(2, -3)$ and $(4, -5),$ then $(a, b)$ is:
  • A
    $(1, 1)$
  • B
    $(-1, 1)$
  • C
    $(1, -1)$
  • ✓
    $(-1, -1)$
Answer
Correct option: D.
$(-1, -1)$
Equation of line passing through the points $(2, -3)$ and $(4, -5)$ is
$\text{y}+3=\frac{-5+3}{4-2}(\text{x}-2)$
$\Rightarrow \text{y}+3=\frac{-2}{2}(\text{x}-2)$
$\Rightarrow \text{y}+3=-(\text{x}-2)$
$\Rightarrow \text{y}+3=-\text{x}+2$
$\Rightarrow \text{x}+\text{y}=-1$
$\Rightarrow \frac{\text{x}}{-1}+\frac{\text{y}}{-1}=1 ($intercept from$)$
$\therefore \text{a}=-1,\text{b}=-1$
Hence, the correct option is $(d).$
View full question & answer→
MCQ 501 Mark
The points $(-a, -b), (0 , 0), (a, b)$ and $(a^2, ab)$ are:
  • A
    Vertices of a square
  • B
    Vertices of a parallelogram
  • ✓
    Collinear
  • D
    Vertices of a rectangle
Answer
Correct option: C.
Collinear
View full question & answer→
MCQ 511 Mark
Distance between the lines $5x + 3y - 7 = 0$ and $15x + 9y + 14 = 0$ is:
  • A
    $\frac{35}{\sqrt{34}}$
  • B
    $\frac{1}{3\sqrt{34}}$
  • ✓
    $\frac{35}{3\sqrt{34}}$
  • D
    $\frac{35}{2\sqrt{34}}$
Answer
Correct option: C.
$\frac{35}{3\sqrt{34}}$
The given lines can be written as
$5\text{x}+3\text{y}-7=0 \ ...(1)$
$5\text{x}+3\text{y}+\frac{14}{3}=0 \ ...(2)$
Let $d$ be the distance between the lines $5x + 3y - 7 = 0$ and $15x + 9y + 14 = 0$
Then, $\text{d}=\Bigg|\frac{-7-\frac{14}{3}}{\sqrt{5^2+3^2}}\Bigg|$
$\Rightarrow\text{d}=\frac{35}{3\sqrt{34}}$
View full question & answer→
MCQ 521 Mark
Choose the correct answer. For specifying a straight line, how many geometrical parameters should be known?
  • A
    $1$
  • ✓
    $2$
  • C
    $4$
  • D
    $3$
Answer
Correct option: B.
$2$
Different form of equation of straight line are slope intercept form, $y = mx + c,$ Paramerer $= 2$
Intercept form, $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$ parameter $= 2$
One-point from, $y - y_1 = m(x - x_1)$, parameter = 2
Normal form, $\text{x}\cos \text{w}+\text{y}\sin\text{w}=\text{P},$ Parameter $= 2$
Hence, the correct option is $(b).$
View full question & answer→
MCQ 531 Mark
A line passes through $P (1, 2)$ such that its intercept between the axes is bisected at $P.$ The equation of the line is:
  • A
    $x + 2y = 5$
  • B
    $x - y + 1 = 0$
  • C
    $x + y - 3 = 0$
  • ✓
    $2x + y - 4 = 0$
Answer
Correct option: D.
$2x + y - 4 = 0$
View full question & answer→
MCQ 541 Mark
Find the equation of line parallel to $x-$axis and passing through $(3, 4):$
  • A
    $x = 3$
  • B
    $x = 4$
  • ✓
    $y = 4$
  • D
    $y = 3$
Answer
Correct option: C.
$y = 4$
Let general equation of line be $y = m \times x + c.$
Since line is parallel to $x-$axis so, $m = 0.$
$\Rightarrow y = c$
$\Rightarrow y = 4$ by substituting the point $(3, 4).$
View full question & answer→
MCQ 551 Mark
If $-40^\circ F$ is equal to $-40^\circ C$ and $0^\circ C$ is equal to $32^\circ F$ then find the value of $40^\circ C:$
  • ✓
    $104^\circ F$
  • B
    $112^\circ F$
  • C
    $86^\circ F$
  • D
    $92^\circ F$
Answer
Correct option: A.
$104^\circ F$
Let general equation be $F = m \times c + k$
$-40 = -40m + k$
and $32 = 0 + k$
$\Rightarrow -40 = -40m + 32$
$\text{ m}=\frac{72}{40} = \frac{18}{10}$
$\text{F}=\frac{18}{10} \times 40 + 32$
$= 72 + 32$
$= 104.$
View full question & answer→
MCQ 561 Mark
The area of a triangle with vertices at $(-4, -1), (1, 2)$ and $(4, -3)$ is:
  • ✓
    $17$
  • B
    $16$
  • C
    $15$
  • D
    None of these.
Answer
Correct option: A.
$17$
Let $A$ be the area of the triangle formed by the points $(-4, -1), (1, 2)$ and $(4, -3).$
$\therefore\text{A}=\frac{1}{2}\big|\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\}\big|$
$\Rightarrow\text{A}=\frac{1}{2}\big|\{-4(2+3)+1(-3+1)+4(-1-2)\}\big|$
$\Rightarrow\text{A}=17$
View full question & answer→
MCQ 571 Mark
If equation of line is $y = 5x + 10$ then find the value of $x-$intercept made by the line:
  • A
    $2$
  • B
    $\frac{1}{2}$
  • C
    $\frac{-1}{2}$
  • ✓
    $-2$
Answer
Correct option: D.
$-2$
Given, equation is $y = 5x + 10.$
$X-$intercept means value of $x$ when $y$ is zero $0 = 5x + 10$
$\Rightarrow x = -2$
View full question & answer→
MCQ 581 Mark
The distance between $M (-1, 5)$ and $N (x, 5)$ is $8$ units. The value of $x$ is:
  • A
    $-9$ or $9$
  • B
    $-7$ or $9$
  • ✓
    $-9$ or $7$
  • D
    $-7$ or $-9$
Answer
Correct option: C.
$-9$ or $7$
$\sqrt{[\text{x}-(-1)^{2}] + (5-5)^{2}} = 8$
$\Rightarrow(\text{x+1})^2=8^2$
$\Rightarrow\text{x}+1=\underline{+}8$
$\therefore\text{x}=-9,7$
View full question & answer→
MCQ 591 Mark
The equation of the straight line which passes through the point $(-4, 3)$ such that the portion of the line between the axes is divided internally by the point in the ratio $5 : 3$ is:
  • ✓
    $9x - 20y + 96 = 0$
  • B
    $9x + 20y = 24$
  • C
    $20x + 9y + 53 = 0$
  • D
    none of these.
Answer
Correct option: A.
$9x - 20y + 96 = 0$
Let the required line intersects the coordinate axis at $(a, 0)$ and $(0, b).$

The point $(−4, 3)$ divides the required line in the ratio $5 : 3$
$\therefore \ -4=\frac{5\times0+3\times\text{a}}{5+3}$ and $3=\frac{5\times\text{b}+3\times0}{5+3}$
$\Rightarrow\text{a}=\frac{ -32}{3}$ and $\text{b}=\frac{ 24}{5}$
Hence, The equation of the required line is given below:
$\frac{\text{x}}{\frac{-32}{3}}+\frac{\text{y}}{\frac{24}{5}}=1$
$\Rightarrow\frac{-3\text{x}}{32}+\frac{5\text{y}}{24}=1$
$\Rightarrow-9\text{x}+20\text{y}=96$
$\Rightarrow9\text{x}-20\text{y}+96=0$
View full question & answer→
MCQ 601 Mark
The slope of a line $ax + by + c = 0$ is:
  • A
    $\frac{\text{a}}{\text{b}}$
  • ✓
    $\frac{\text{-a}}{\text{b}}$
  • C
    $\frac{\text{c}}{\text{b}}$
  • D
    $\frac{\text{-c}}{\text{b}}$
Answer
Correct option: B.
$\frac{\text{-a}}{\text{b}}$
We know that the general equation of a line is $ax + by + c = 0.$
Rearranging the equation, we get
$\Rightarrow by = -ax - c$
$\Rightarrow\text{y} =\big(\frac{\text{-a}}{\text{b}})\text{ x}-\big(\frac{\text{-c}}{\text{b}}) ...(1)$
This is of the form, $y = mx + c … (2)$
By comparing $(1)$ and $(2),$ we get
Slope, $\text{m} = \frac{\text{-a}}{\text{b}}$
View full question & answer→
MCQ 611 Mark
Two vertices of a triangle are $(-2, -1)$ and $(3, 2)$ and third vertex lies on the line $x + y = 5.$ If the area of the triangle is $4$ square units, then the third vertex is:
  • A
    $(0, 5)$ or, $(4, 1)$
  • ✓
    $(5, 0)$ or, $(1, 4)$
  • C
    $(5, 0)$ or, $(4, 1)$
  • D
    $(0, 5)$ or, $(1, 4)$
Answer
Correct option: B.
$(5, 0)$ or, $(1, 4)$
Let $(h, k)$ be the third vertex of the triangle.
It is given that the area of the triangle with vertices $(h, k), (-2, -1)$ and $(3, 2)$ is $4$ square units.
$\frac{1}{2}\big|\text{h}(-1-2)-3(-1-\text{K})-2(2-\text{K})\big|=4$
$\Rightarrow3\text{h}-5\text{k}+1=\pm 8$
Taking positive sign, we get,
$3h - 5k + 1 = 8$
$3h - 5k - 7 = 0 ...(1)$
Taking negative sign, we get,
$3h - 5k + 9 = 0 ...(2)$
The vertex $(h, k)$ lies on the line $x + y = 5.$
$h + k - 5 = 0 ...(3)$
On solving $(1)$ and $(3)$, we find $(4, 1)$ to be the coordinates of the third vertex.
Similarly, on solving $(2)$ and $(3),$ we find $(2, 3)$ to be the coordinates of the third vertex.
View full question & answer→
MCQ 621 Mark
If $x-$intercept of a line is $4$ and its $y-$intercept is $2$ then find the equation of line:
  • A
    $2x + y - 4 = 0.$
  • ✓
    $x + 2y - 4 = 0.$
  • C
    $2x + y + 4 = 0.$
  • D
    $x + 2y + 4 = 0.$
Answer
Correct option: B.
$x + 2y - 4 = 0.$
If $x-$intercept of a line is a and $y-$intercept of line is b so, equation of line is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}} = 1$
Equation of line is $\frac{\text{x}}{4}+\frac{\text{y}}{2} = 1$
$\Rightarrow x + 2y - 4 = 0.$
View full question & answer→
MCQ 631 Mark
Find the distance between $2x + y + 4 = 0$ and $2x + y + 8 = 0:$
  • ✓
    $\frac{4}{\sqrt5}$
  • B
    $\frac{3}{\sqrt5}$
  • C
    $\frac{9}{\sqrt5}$
  • D
    $\frac{3}{\sqrt5}$
Answer
Correct option: A.
$\frac{4}{\sqrt5}$
Distance between parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is $\mid\frac{\text{c}_{1}-\text{c}_{2}}{\sqrt{\text{a}^2+\text{b}^2}}\mid$
So, distance $2x + y + 4 = 0$ and $2x + y + 8 = 0$ is $\mid\frac{8-4}{\sqrt2^2+1^2}\mid=\frac{4}{\sqrt5}$
View full question & answer→
MCQ 641 Mark
If equation of a line is $y = 3x - 4$ then find the slope of line:
  • ✓
    $3$
  • B
    $-3$
  • C
    $4$
  • D
    $-4$
Answer
Correct option: A.
$3$
Comparing the above equation with general equation $y = m × x + c,$
$m = 3$ which is the slope of line.
View full question & answer→
MCQ 651 Mark
Choose the correct answer. The distance between the lines $y = mx + c_1$ and $y = mx + c_2$ is:
  • A
    $\frac{\text{c}_1-\text{c}_2}{\sqrt{\text{m}^2+1}}$
  • ✓
    $\frac{|\text{c}_1-\text{c}_2|}{\sqrt{1+\text{m}^2}}$
  • C
    $\frac{\text{c}_2-\text{c}_1}{\sqrt{1+\text{m}^2}}$
  • D
    $0$
Answer
Correct option: B.
$\frac{|\text{c}_1-\text{c}_2|}{\sqrt{1+\text{m}^2}}$
Let any point on the line $y=m x+c_1$ be $P\left(x_1, y_1\right)$.
The equation of the other line is : $y=m x+c_2$
$\Rightarrow m x-y+c_2=0$
Distance of point $P$ from this line, $\text{d}=\frac{|\text{mx}_1-\text{y}_1+\text{c}_2|}{\sqrt{\text{m}^2+1}}$
Since $P$ line on the first line, we get
$ \Rightarrow y_1=m x_1+c_1$
$ \Rightarrow m x_1-y_1=-c_1 $
$\therefore \text{d}=\frac{|\text{c}_1-\text{c}_2|}{\sqrt{\text{m}^2+1}}$
View full question & answer→
MCQ 661 Mark
Angle made by line with measured anticlockwise is called inclination of the line:
  • ✓
    Positive $x-$axis
  • B
    Negative $x-$axis
  • C
    Positive $y-$axis
  • D
    Negative $y-$axis
Answer
Correct option: A.
Positive $x-$axis
We know, inclination of line is always measured with positive $x-$axis in anticlockwise direction.
View full question & answer→
MCQ 671 Mark
Choose the correct answer.
If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is $(3, 2),$ then the equation of the line will be:
  • ✓
    $2x + 3y = 12$
  • B
    $3x + 2y = 12$
  • C
    $4x - 3y = 6$
  • D
    $5x - 2y = 10$
Answer
Correct option: A.
$2x + 3y = 12$
Let the given the line meets the axes at $A(a, 0)$ and $B(0, b).$
Given that $C(3, 2)$ is the mid$-$point of $AB$
$\therefore 3=\frac{\text{a}+0}{2}\Rightarrow \text{a}=6$
and $2=\frac{0+\text{b}}{2}\Rightarrow \text{b}=4$

Intercept form of the line $AB$
$\Rightarrow \frac{\text{x}}{6}+\frac{\text{y}}{4}=1$
$\Rightarrow 2\text{x}+3\text{y}=12$
Hence, the correct option is $(a).$
View full question & answer→
MCQ 681 Mark
If slope of a line is negative then its inclination is:
  • A
    Right angle.
  • B
    Acute angle.
  • ✓
    Obtuse angle.
  • D
    Zero.
Answer
Correct option: C.
Obtuse angle.
If inclination is $\alpha$ slope is given by $\tan\alpha$ Given that slope of line is negative which means $\tan\alpha$ is negative. We know, $\tan\alpha$ is negative in $2^{nd}$ quadrant
i.e. $\alpha$ should be obtuse angle.
View full question & answer→
MCQ 691 Mark
Choose the correct answer. Equations of diagonals of the square formed by the lines $x = 0, y = 0, x = 1$ and $y = 1$ are:
  • ✓
    $y = x, y + x = 1$
  • B
    $y = x, x + y = 2$
  • C
    $2y = x, y + x = \frac{1}{3}$
  • D
    $y = 2x, y + 2x = 1$
Answer
Correct option: A.
$y = x, y + x = 1$
Given lines are plotted on coordinate plane as shown in the adjacent figure.
From the figure, equation of diagonal $OB$ is $y = x.$
Equation of the diagonal $AC$ is $x + y = 1 ($using intercept form$).$
View full question & answer→
MCQ 701 Mark
If $p$ be the length of the perpendicular from the origin on the line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$ then:
  • A
    $\text{p}^2=\text{a}^2+\text{b}^2$
  • B
    $\text{p}^2=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
  • ✓
    $\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
  • D
    None of these.
Answer
Correct option: C.
$\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
It is given that $p$ is the length of the perpendicular from the origin on the line
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
$\frac{1}{\text{a}}\text{x}+\frac{1}{\text{b}}\text{y}-1=0$
$\therefore\text{p}=\begin{vmatrix}\frac{0+0+1}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}}} \end{vmatrix}$
Squaring both sides,
$\Rightarrow\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
View full question & answer→
MCQ 711 Mark
If equation of line is $y = 5x + 10$ then find the value of $x-$intercept made by the line:
  • A
    $2$
  • B
    $\frac{1}{2}$
  • C
    $\frac{-1}{2}$
  • ✓
    $-2$
Answer
Correct option: D.
$-2$
Given, equation is $y = 5x + 10.$
$X-$intercept means value of $x$ when $y$ is zero.
​​​​​​​$0 = 5x + 10$
$\Rightarrow x = -2$
View full question & answer→
MCQ 721 Mark
Slope of a line is given by if inclination of line is $\alpha$:
  • A
    $\sin\alpha$
  • B
    $\cos\alpha$
  • ✓
    $\tan\alpha$
  • D
    $\cot\alpha$
Answer
Correct option: C.
$\tan\alpha$
Slope of a line is given by $\tan\alpha$ if inclination of line is $\alpha$ Slope is denoted by tangent of the inclination angle.
View full question & answer→
MCQ 731 Mark
The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between the line $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is:
  • A
    $1 : 2$
  • ✓
    $3 : 7$
  • C
    $2 : 3$
  • D
    $2 : 5$
Answer
Correct option: B.
$3 : 7$
Here, in all equations the coefficient of $x$ is same.
It means all the lines have same slope
So, all the lines are parallel.
Now, the distance between the line $3x + 4y + 2 = 0$ and $3x + 4y + 5 = 0$ is given by
$\frac{|2-5|}{\sqrt{3^2+4^2}}$
$=\frac{3}{\sqrt{25}}=\frac{3}{5}$
Hence, the ratio is given by
$\frac{3}{5}:\frac{7}{5}$
$=3:7$
View full question & answer→
MCQ 741 Mark
If a line makes an angle a with the positive direction of $x-$axis, then the slope of the line is given by:
  • A
    $\text{m} = \sin\text{a}$
  • B
    $\text{m} = \cos\text{a}$
  • ✓
    $\text{m} = \tan\text{a}$
  • D
    $\text{m} = \sec\text{a}$
Answer
Correct option: C.
$\text{m} = \tan\text{a}$
View full question & answer→
MCQ 751 Mark
The equation of a straight line that passes through the point $(3, 4)$ and perpendicular to the line $3x + 2y + 5 = 0$ is:
  • ✓
    $2x - 3y + 6 = 0$
  • B
    $2x + 3y + 6 = 0$
  • C
    $2x - 3y - 6 = 0$
  • D
    $2x + 3y - 6 = 0$
Answer
Correct option: A.
$2x - 3y + 6 = 0$
The equation of a straight line perpendicular to $3x + 2y + 5 = 0$ is $2\text{x}-3\text{y}+\lambda=0 …(1)$
This passes through the point $(3, 4).$
Now, substitute in equation $(1),$ we get
$2(2) - 3(4) +\lambda = 0$
$4-12+\lambda =0$
$-6+\lambda =0$
$\lambda=6$
Substituting $\lambda=6 $ in $(1),$ we get $2x - 3y + 6 = 0,$ which is the required equation.
View full question & answer→
MCQ 761 Mark
The inclination of the straight line passing through the point $(-3, 6)$ and the mid$-$point of the line joining the point $(4, -5)$ and $(-2, 9)$ is:
  • A
    $\frac{\pi}{4}$
  • B
    $\frac{\pi}{6}$
  • C
    $\frac{\pi}{3}$
  • ✓
    $\frac{3\pi}{4}$
Answer
Correct option: D.
$\frac{3\pi}{4}$
The midpoint of the line joining the points $(4, -5)$ and $(-2, 9)$ is $(1, 2).$
Let $\theta$ be the inclination of the straight line passing through the points $(-3, 6)$ and $(1, 2).$
Then, $\tan\theta=\frac{ 2-6}{1+3}=-1$
$\Rightarrow\theta=\frac{3\pi}{4}$
View full question & answer→
MCQ 771 Mark
If slope of a line is $4$ and $x-$intercept made by the line is $2$ then the equation of line will be:
  • ✓
    $y = 4x - 8$
  • B
    $y = 4x + 8$
  • C
    $y = 2x + 4$
  • D
    $y = 2x - 4$
Answer
Correct option: A.
$y = 4x - 8$
Let general equation of line be $y = m \times x + c.$
Given, $m = 4$ and value of $x$ when $y = 0$ is $2.$
$C = -m \times 2 = -4 \times 2 = -8.$
$\Rightarrow y = 4x - 8$
View full question & answer→
MCQ 781 Mark
The angle between the lines $2x - y + 3 = 0$ and $x + 2y + 3 = 0$ is:
  • ✓
    $90^\circ$
  • B
    $60^\circ$
  • C
    $45^\circ$
  • D
    $30^\circ$
Answer
Correct option: A.
$90^\circ$
Let $m_1$ and $m_2$ be the slope of the lines $2x - y + 3 = 0$ and $x + 2y + 3 = 0,$ respectively.
Let $\theta$ be the angle between them.
Here, $m_1= 2$ and $\text{m}_2=-\frac{1}{2}$
$\because\text{m}_1\text{m}_2=-1$
Therefore, the angle between the given lines is $90^\circ .$
View full question & answer→
MCQ 791 Mark
If $A (6, 4)$ and $B (2, 12)$ are the two points, then the slope of a line perpendicular to line $AB$ is:
  • A
    $-2$
  • B
    $2$
  • ✓
    $\frac{1}{2}$
  • D
    $\frac{-1}{2}$
Answer
Correct option: C.
$\frac{1}{2}$
Given points: $A (6, 4) =(x_1, y_1)$
$B (2, 12) =(x_2, y_2)$
We know that the slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\frac{(\text{y}_{2}\text{-y}_{1})}{(\text{x}_{2}-\text{x}_{1})}$
$\text{m} = \frac{(12-4)}{(2-6)} = \frac{8}{-4} = -2.$
We know that the slope of two perpendicular lines $m_1.m_2 =-1.$
The slope of a line perpendicular to line $AB$ is $\frac{-1}{\text{m}} = \frac{-1}{-2} = \frac{1}{2}$
View full question & answer→
MCQ 801 Mark
$A(6, 3), B(-3, 5), C(4, -2)$ and $D(x, 3x)$ are four points. If $\triangle\text{DBC} : \triangle\text{ABC}= 1 : 2,$ then $x$ is equal to:
  • ✓
    $\frac{11}{8}$
  • B
    $\frac{8}{11}$
  • C
    $3$
  • D
    None of these
Answer
Correct option: A.
$\frac{11}{8}$
The area of a triangle with vertices $D(x, 3x), B(-3, 5)$ and $C(4, -2)$ is given below:
Area of $\triangle\text{DBC}=\frac{1}{2}\{\text{x}(5+2)-3(-2-3\text{x})+4(3\text{x}-5)\}$
$\Rightarrow$ Area of $\triangle\text{DBC}=(14\text{x}-7)\text{sq units}$
Similarly, the area of a triangle with vertices $A(6, 3), B(-3, 5)$ and $C(4, -2)$ is given below:
$\triangle\text{ABC}=\frac{1}{2}\{6(5+2)-3(-2-3)+4(3-5)\}$
$\triangle\text{ABC}=\frac{49}{2}\text{sq units}$
Given:
$\triangle\text{DBC}:\triangle\text{ABC}=1:2$
$\frac{2(14\text{x}-7)}{49}=\frac{1}{2}$
$\Rightarrow8\text{x}-4=7$
$\Rightarrow\text{x}=\frac{11}{8}$
View full question & answer→
MCQ 811 Mark
The equation $\frac{(2 \text{x})}{2} - \frac{(\text{y}{ 2})}{2} = -3,$ if $x = -2,$ then $y$ is equal to:
  • A
    $-2$
  • B
    $-7$
  • ✓
    $8$
  • D
    $5$
Answer
Correct option: C.
$8$
View full question & answer→
MCQ 821 Mark
Equation of vertical line to the left of $y-$axis at $5$ units from $y-$axis is:
  • A
    $x = 5$
  • ✓
    $x = -5$
  • C
    $y = 5$
  • D
    $y = -5$
Answer
Correct option: B.
$x = -5$
Equation of $y-$axis is $x = 0.$ Vertical line is parallel to $y-$axis and to the left by $5$ units
so, equation of line is $x = -5.$
View full question & answer→
MCQ 831 Mark
Find the equation perpendicular to $2x - y = 4$ and pass through $(2, 4):$
  • A
    $2x + y - 10 = 0$
  • B
    $x + 2y + 10 = 0$
  • ✓
    $x + 2y - 10 = 0$
  • D
    $x + y - 10 = 0$
Answer
Correct option: C.
$x + 2y - 10 = 0$
Line $2x - y = 4$ has slope $2$.
Line perpendicular to given line has slope $\frac{-1}{2}$
equation is $(\text{y}-4) = \big(\frac{-1}{2}\big) (\text{x}-2)$
$2y - 8 = -x + 2$
$\Rightarrow x + 2y - 10 = 0$
View full question & answer→
MCQ 841 Mark
Which of the following lines is farthest from the origin?
  • A
    $x - y + 1 = 0$
  • B
    $2x - y + 3 = 0$
  • C
    $x + 2y - 2 = 0$
  • ✓
    $x + y - 2 = 0$
Answer
Correct option: D.
$x + y - 2 = 0$
View full question & answer→
MCQ 851 Mark
Equation of horizontal line above $x-$axis at $5$ units from $x-$axis is:
  • A
    $x = 5$
  • B
    $x = -5$
  • ✓
    $y = 5$
  • D
    $y = -5$
Answer
Correct option: C.
$y = 5$
Equation of $x-$axis is $y = 0.$ Horizontal line is parallel to $x-$axis and above it by $5$ units
so, equation of line is $y = 5.$
View full question & answer→
MCQ 861 Mark
The inclination of the line $x - y + 3 = 0$ with the positive direction of $x-$axis is:
  • ✓
    $45^\circ$
  • B
    $135^\circ$
  • C
    $-45^\circ$
  • D
    $-135^\circ$
Answer
Correct option: A.
$45^\circ$
View full question & answer→
MCQ 871 Mark
What is the inclination of a line which is parallel to $y-$axis?
  • A
    $0^\circ$
  • B
    $180^\circ$
  • C
    $45^\circ$
  • ✓
    $90^\circ$
Answer
Correct option: D.
$90^\circ$
If a line is parallel to $y-$axis then angle formed by it with $x-$axis is $90^\circ .$
So, its inclination is $90^\circ .$
View full question & answer→
MCQ 881 Mark
Find the equation of line parallel to $x-$axis and passing through $(3, 4):$
  • A
    $x = 3$
  • B
    $x = 4$
  • ✓
    $y = 4$
  • D
    $y = 3$
Answer
Correct option: C.
$y = 4$
Let general equation of line be $y = m \times x + c.$
Since line is parallel to $x-$axis so, $m = 0.$
$\Rightarrow y = c$
$\Rightarrow y = 4$ by substituting the point $(3, 4).$
View full question & answer→
MCQ 891 Mark
Equation of horizontal line above $x-$axis at $5$ units from $x-$axis is:
  • A
    $x = 5$
  • B
    $x = -5$
  • ✓
    $y = 5$
  • D
    $y = -5$
Answer
Correct option: C.
$y = 5$
Equation of $x-$axis is $y = 0.$ Horizontal line is parallel to $x-$axis and above it by $5$ units
so, equation of line is $y = 5$
View full question & answer→
MCQ 901 Mark
The equations of the sides $\text{AB, BC}$ and $\text{CA}$ of $\triangle \text{ABC}$ are $y - x = 2, x + 2y = 1$ and $3x + y + 5 = 0$ respectively. The equation of the altitude through $B$ is:
  • A
    $x - 3y + 1 = 0$
  • ✓
    $x - 3y + 4 = 0$
  • C
    $3x - y + 2 = 0$
  • D
    None of these.
Answer
Correct option: B.
$x - 3y + 4 = 0$
The equation of the sides $\text{AB, BC}$ and $\text{CA}$ of $\triangle \text{ABC}$ are $y - x = 2, x + 2y = 1$ and $3x + y + 5 = 0,$ respectively.
Solving the equations of $\text{AB}$ and $\text{BC,}$
i.e. $y - x = 2$ and $x + 2y = 1,$ we get:
$x = -1, y = 1$
So, the coordinates of $B$ are $(-1, 1).$
The altitude through $B$ is perpendicular to $AC.$
$\therefore$ Slope of $AC = -3$
Thus, slope of the altitude through $B$ is $13.$
Thus, slope of the altitude through $B$ is $\frac{1}{3}.$
Equation of the required altitude is given below:
$\text{y}-1+\frac{1}{3}(\text{x}+1)$
$\Rightarrow\text{x}-3\text{y}+4=0$
View full question & answer→
MCQ 911 Mark
For specifying a straight line, how many geometrical parameters should be known:
  • A
    $1$
  • ✓
    $2$
  • C
    $4$
  • D
    $3$
Answer
Correct option: B.
$2$
View full question & answer→
MCQ 921 Mark
Two lines are said to be parallel if the difference of their slope is:
  • A
    $-1$
  • ✓
    $0$
  • C
    $1$
  • D
    None of these
Answer
Correct option: B.
$0$
We know that two lines are said to be parallel if their slope is equal. If $m_1$ and $m_2$ are the slopes of two parallel lines, then it is represented as $m_1= m_2$.
The difference of their slope should be $m_1- m_2= 0.$
View full question & answer→
MCQ 931 Mark
The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is:
  • A
    $\frac{3}{2}$
  • ✓
    $\frac{3}{10}$
  • C
    $6$
  • D
    $\frac{9}{4}$
Answer
Correct option: B.
$\frac{3}{10}$
View full question & answer→
MCQ 941 Mark
If equation of line is $x + y = 2$ then find the angle made by line with $x-$axis:
  • ✓
    $45^\circ$
  • B
    $60^\circ$
  • C
    $30^\circ$
  • D
    $75^\circ$
Answer
Correct option: A.
$45^\circ$
Given, equation is $x + y = 2.$ Reducing the above equation to normal form
$\frac{(\text{x + y})}{\sqrt2} ={\sqrt2}.$
$\text{x} \cos⁡45^\circ + \text{y} \sin 45^\circ = {\sqrt2}$
Angle made with $x-$axis is $45^\circ .$
View full question & answer→
MCQ 951 Mark
Angle made by line with measured anticlockwise is called inclination of the line:
  • ✓
    Positive $x-$axis.
  • B
    Negative $x-$axis.
  • C
    Positive $y-$axis.
  • D
    Negative $y-$axis.
Answer
Correct option: A.
Positive $x-$axis.
We know, inclination of line is always measured with positive $x-$axis in anticlockwise direction.
View full question & answer→
MCQ 961 Mark
If $a + b + c = 0,$ then the family of lines $3ax + by + 2c = 0$ pass through fixed point:
  • A
    $\Big(2,\frac{2}{3}\Big)$
  • ✓
    $\Big(\frac{2}{3},2\Big).$
  • C
    $\Big(-2,\frac{2}{3}\Big)$
  • D
    None of these.
Answer
Correct option: B.
$\Big(\frac{2}{3},2\Big).$
Given:
$a + b + c = 0$
Substituting $c = -a - b$ in $3ax + by + 2c = 0,$ we get:
$3\text{ax}+\text{by}-2\text{a}-2\text{b}=0$
$\Rightarrow\text{a}(3\text{x}-2)+\text{b}(\text{y}-2)=0$
$\Rightarrow(3\text{x}-2)+\frac{\text{b}}{\text{a}}(\text{y}-2)=0$
This line is of the form $\text{L}_1+\lambda\text{L}_2=0,$ which passes through the intersection of the lines $L_1$ and $L_2$ 
i.e. $3x - 2 = 0$ and $y - 2 = 0.$
Solving $3x - 2 = 0$ and $y - 2 = 0,$ we get
$\text{x}=\frac{2}{3},\text{y}=2$
Hence, the required fixed point is $\Big(\frac{2}{3},2\Big).$
View full question & answer→
MCQ 971 Mark
Equation of the straight line making equal intercepts on the axes and passing through the point $(2, 4)$ is:
  • A
    $4x - y - 4 = 0$
  • B
    $2x + y - 8 = 0$
  • ✓
    $x + y - 6 = 0$
  • D
    $x + 2y - 10 = 0$
Answer
Correct option: C.
$x + y - 6 = 0$
View full question & answer→
MCQ 981 Mark
Equation of vertical line to the right of $y-$axis at $5$ units from $y-$axis is:
  • ✓
    $x = 5$
  • B
    $x = -5$
  • C
    $y = 5$
  • D
    $y = -5$
Answer
Correct option: A.
$x = 5$
Equation of $y-$axis is $x = 0.$ Vertical line is parallel to $y-$axis and to the right by $5$ units
so, equation of line is $x = 5.$
View full question & answer→
MCQ 991 Mark
What can be said regarding a line if its slope is negative?
  • A
    $\theta$ is an acute angle
  • ✓
    $\theta$ is an obtuse angle
  • C
    Either the line is $x-$axis or it is parallel to the $x-$axis
  • D
    None of these
Answer
Correct option: B.
$\theta$ is an obtuse angle
The line with a negative slope makes an obtuse angle with a positive $x-$axis when measured in the anti$-$clockwise direction.
View full question & answer→
MCQ 1001 Mark
The centroid of the triangle with vertices $(2, 6), (-5, 6)$ and $(9,3)$ is:
  • A
    $(2, -3)$
  • ✓
    $(2, 5)$
  • C
    $(-2, 3)$
  • D
    $(-2, -3)$
Answer
Correct option: B.
$(2, 5)$
$\text{G}=\Big(\frac{\text{x}_{1}+\text{x}_{2}+\text{x}_{3}}{3},\frac{\text{y}_{1}+\text{y}_{2}+\text{y}_{3}}{3}\Big)$
$=\Big(\frac{2 - 5 + 9}{3},\frac{6 + 6 + 6}{3}\Big)$
$= (2,5)$
View full question & answer→
MCQ 1011 Mark
If a line has slope $3$ and pass through point $(1, 2)$ then the equation of line is:
  • A
    $x = 3y - 1$
  • B
    $x = 3y + 1$
  • C
    $y = 3x + 1$
  • ✓
    $y = 3x - 1$
Answer
Correct option: D.
$y = 3x - 1$
Let general equation of line be $y = m \times x + c.$
Given $m = 3$
$\Rightarrow y = 3x + c$
Substituting the point $(1, 2)$ in above equation we get $2 = 3 \times 1 + c$
$\Rightarrow c = -1$
So, equation of line will be $y = 3x - 1.$
View full question & answer→
MCQ 1021 Mark
The inclination of the line $5x - 5y + 8 = 0$ is:
  • A
    $30^\circ$
  • ✓
    $45^\circ$
  • C
    $60^\circ$
  • D
    $90^\circ$
Answer
Correct option: B.
$45^\circ$
View full question & answer→
MCQ 1031 Mark
The equation of the line through the points $(1, 5)$ and $(2, 3)$ is:
  • A
    $2x - y - 7 = 0$
  • B
    $2x + y + 7 = 0$
  • ✓
    $2x + y - 7 = 0$
  • D
    $x + 2y - 7 = 0$
Answer
Correct option: C.
$2x + y - 7 = 0$
View full question & answer→
MCQ 1041 Mark
The centroid of the triangle with vertices $(2, 6), (-5, 6)$ and $(9, 3)$ is:
  • A
    $(2, -3)$
  • ✓
    $(2, 5)$
  • C
    $(-2, 3)$
  • D
    $(-2, -3)$
Answer
Correct option: B.
$(2, 5)$
$\text{G} =\Big(\frac{ {\text{x}_{1}+\text{x}_{2}+\text{x}_{3}}}{3},\frac{ {\text{y}_{1}+\text{y}_{2}+\text{y}_{3}}}{3}\Big)$
$=\Big(\frac {2-5+9}{3}, \frac{6+6+6}{3}\Big)$
$= (2,5)$
View full question & answer→
MCQ 1051 Mark
If the lines $x + q = 0, y - 2 = 0$ and $3x + 2y + 5 = 0$ are concurrent, then the value of $q$ will be:
  • A
    $1$
  • B
    $2$
  • ✓
    $3$
  • D
    $5$
Answer
Correct option: C.
$3$
The lines $x + q = 0, y - 2 = 0$ and $3x + 2y + 5 = 0$ are concurrent.
$\therefore\begin{vmatrix}1&0&\text{q}\\0&1&-2\\3&2&5 \end{vmatrix}=0$
$\Rightarrow1(5+4)-0+\text{q}(0-3)=0$
$\Rightarrow3\text{q}=9$
$\Rightarrow\text{q}=3$
View full question & answer→
MCQ 1061 Mark
The distance of the point $(x_1, y_1)$ from the origin:
  • A
    $\text{x}\frac{2}{1}+\text{y}\frac{2}{1}$
  • ✓
    $\sqrt{\text{x}\frac{2}{1}+\text{y}\frac{2}{1}}$
  • C
    $\frac{1}{\sqrt{\text{x}\frac{2}{1}+\text{y}\frac{2}{1}}}$
  • D
    $\frac{1}{\text{x}\frac{2}{1}+\text{y}\frac{2}{1}}$
Answer
Correct option: B.
$\sqrt{\text{x}\frac{2}{1}+\text{y}\frac{2}{1}}$
We can use the distance formula to find the length of the line.
Distance between points $\sqrt{(\text{x}_{2}-\text{x}_{1}+(\text{y}_{2}-\text{y}_{1})}$
But given that one of the co$-$ordinates is $(0, 0)$
Distance from $(0, 0)$ is $=\sqrt{(\text{x}_{1})^{2}+(\text{y}_{1})^{2}}$
View full question & answer→
MCQ 1071 Mark
A point equidistant from the line $4x + 3y + 10 = 0, 5x - 12y + 26 = 0$ and $7x + 24y - 50 = 0$ is:
  • A
    $(1, -1)$
  • B
    $(1, 1)$
  • ✓
    $(0, 0)$
  • D
    $(0, 1)$
Answer
Correct option: C.
$(0, 0)$
Let the coordiantes of the point be $(a, b)$
Now, the distance of the point $(a, b)$ from $4x + 3y + 10 = 0$ is given by
$\Bigg|\frac{4\text{a}+3\text{b}+10}{\sqrt{4^2+3^2}}\Bigg|$
$=\Bigg|\frac{4\text{a}+3\text{b}+10}{5}\Bigg|$
Again, the distance of the point $(a, b)$ from $5x - 12y + 26 = 0$ is given by
$\Bigg|\frac{5\text{a}-12\text{b}+26}{\sqrt{5^2+(-12)^2}}\Bigg|$
$=\Bigg|\frac{5\text{a}-12\text{b}+26}{13}\Bigg|$
Again, the distance of the point $(a, b)$ from $7x + 24y - 50 = 0$ is is given by
$\Bigg|\frac{7\text{a}+24\text{b}-50}{\sqrt{7^2+(24)^2}}\Bigg|$
$=\Bigg|\frac{7\text{a}+24\text{b}-50}{25}\Bigg|$
Now,
$\Bigg|\frac{4\text{a}+3\text{b}+10}{5}\Bigg|=\Bigg|\frac{5\text{a}-12\text{b}+26}{13}\Bigg|=\Bigg|\frac{7\text{a}+24\text{b}-50}{25}\Bigg|$
Only $a = 0$ and $b = 0$ is satisfying the above equation
Hence, the correct answer is option $(c).$
View full question & answer→
MCQ 1081 Mark
Equation of the line passing through $(0, 0)$ and slope $m$ is:
  • A
    $y = mx + c$
  • B
    $x = my + c$
  • ✓
    $y = mx$
  • D
    $x = my$
Answer
Correct option: C.
$y = mx$
View full question & answer→
MCQ 1091 Mark
Area of the triangle formed by the points $((a + 3)(a + 4), a + 3), ((a + 2)(a + 3), (a + 2))$ and $((a + 1)(a + 2), (a + 1))$ is:
  • A
    $25a^2$
  • B
    $5a^2$
  • C
    $24a^2$
  • ✓
    None of these.
Answer
Correct option: D.
None of these.
The given points are $\{(a + 3)(a + 4), a + 3), ((a + 2)(a + 3), (a + 2))$ and $((a + 1)(a + 2), (a + 1)\}$
Let $A$ be the area of the triangle formed by these points.
Then, $\text{A}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$\Rightarrow\text{A}=\frac{1}{2}[(\text{a}+3)(\text{a}+4)(\text{a}+2-\text{a}-1)+(\text{a}+2)(\text{a}+3)\text{a}+1-\text{a}-3)+(\text{a}+1)(\text{a}+2)(\text{a}+3-\text{a}-2)]$
$\Rightarrow\text{A}=\frac{1}{2}[(\text{a}+3)(\text{a}+4)-2(\text{a}+2)(\text{a}+3)+(\text{a}+1)(\text{a}+2)]$
$\Rightarrow\text{A}=\frac{1}{2}[\text{a}^2+7\text{a}+12-2\text{a}^2-10\text{a}-12+\text{a}^2+3\text{a}+2]$
$\Rightarrow\text{A}=1$
View full question & answer→
MCQ 1101 Mark
If slope of a line is negative then its inclination is:
  • A
    Right angle
  • B
    Acute angle
  • ✓
    Obtuse angle
  • D
    Zero
Answer
Correct option: C.
Obtuse angle
If inclination is α slope is given by $\tan\alpha$ Given that slope of line is negative which means $\tan\alpha$ is negative.
We know, $\tan\alpha$ is negative in $2^{nd}$ quadrant
i.e. $\alpha$ should be obtuse angle
View full question & answer→
MCQ 1111 Mark
Choose the correct answer. The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between the lines $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is:
  • A
    $1 : 2$
  • ✓
    $3 : 7$
  • C
    $2 : 3$
  • D
    $2 : 5$
Answer
Correct option: B.
$3 : 7$
Given lines are:
$3x + 4y + 5 = 0 .....(i)$
$3x + 4y - 5 = 0 .....(ii)$
The third line is : $3x + 4y + 2 = 0$
Distance between the line $(i)$ and $(iii) =\frac{|5-2|}{\sqrt{9+16}}=\frac{3}{5}$
Distance between the lines $(ii)$ and $(iii) =\frac{|-5-2|}{\sqrt{9+16}}=\frac{7}{5}$
Hence, the required ratio is $\frac{3}{5}:\frac{7}{5}$ or $3 : 7.$
View full question & answer→
MCQ 1121 Mark
The area of the triangle formed by the points $(a, b + c), (b, c + a)$ and $(c, a + b)$ is:
  • A
    $1$
  • B
    $a + b + c$
  • C
    $abc$
  • ✓
    $0$
Answer
Correct option: D.
$0$
Area $=\frac{1}{2}[\text{a}(\text{c + a - a - b})+\text{b }(\text{a + b - b - c})+\text{c }(\text{b + c - c - a})]=0$
View full question & answer→
MCQ 1131 Mark
If two vertices of a triangle are $(3, -2)$ and $(-2, 3)$ and its orthocenter is $(-6, 1)$ then its third vertex is:
  • A
    $(5, 3)$
  • B
    $(-5, 3)$
  • C
    $(5, -3)$
  • ✓
    $(-5, -3)$
Answer
Correct option: D.
$(-5, -3)$
View full question & answer→
MCQ 1141 Mark
The equation of the line with slope $-\frac{3}{2}$ and which is concurrent with the lines $4x + 3y - 7 = 0$ and $8x + 5y - 1 = 0$ is:
  • A
    $3x + 2y - 63 = 0$
  • ✓
    $3x + 2y - 2 = 0$
  • C
    $2y - 3x - 2 = 0$
  • D
    None of these.
Answer
Correct option: B.
$3x + 2y - 2 = 0$
Given:
$4x + 3y - 7 = 0 ...(1)$
$8x + 5y - 1 = 0 ...(2)$
The equation of the line with slope $-\frac{3}{2}$ is given below:
$\text{y}=-\frac{3}{2}\text{x}+\text{c}$
$\Rightarrow\frac{3}{2}\text{x}+\text{y}-\text{C}=0 \ ...(3)$
The lines $(1), (2)$ and $(3)$ are concurrent.
$\therefore\begin{vmatrix}4&3&-7\\8&5&-1\\\frac{3}{2}&1&\text{-c} \end{vmatrix}=0$
$\Rightarrow4(-5\text{c}+1)-3\Big(-8\text{c}+\frac{3}{2}\Big)-7\Big(8-\frac{15}{2}\Big)=0$
$\Rightarrow-20\text{c}+4+24\text{c}-\frac{9}{2}-56+\frac{105}{2}=0$
$\Rightarrow\frac{-40\text{c}+8+48\text{c}-9-112+105}{2}=0$
$\Rightarrow8\text{c}=8$
$\Rightarrow\text{c}=1$
On substituting $c = 1$ in $\text{y}=-\frac{3}{2}\text{x}+\text{c},$ we get:
$\text{y}=-\frac{3}{2}\text{x}+1,$
$\Rightarrow3\text{x}+2\text{y}-2=0$
View full question & answer→
MCQ 1151 Mark
Choose the correct answer. The point $(4, 1)$ undergoes the following two successive transformations:
  • A
    $(4, 3)$
  • ✓
    $(3, 4)$
  • C
    $(1, 4)$
  • D
    $\frac{7}{2},\frac{7}{2}$
Answer
Correct option: B.
$(3, 4)$
Reflection of $A(4, 1)$ in $y = x$ is $5(1, 4).$
Now translation of point $B$ through a distance $'2\ '$ units along the positive $x-$axis shifts $B$ to $C(1 + 2, 4)$ or $C(3, 4).$
View full question & answer→
MCQ 1161 Mark
Choose the correct answer. The distance of the point of intersection of the lines $2x - 3y + 5 = 0$ and $3x + 4y = 0$ from the line $5x - 2y = 0$ is:
  • ✓
    $\frac{130}{17\sqrt{29}}$
  • B
    $\frac{13}{7\sqrt{29}}$
  • C
    $\frac{130}{7}$
  • D
    None of these.
Answer
Correct option: A.
$\frac{130}{17\sqrt{29}}$
Given lines are:
$2x - 3y + 5 = 0 .....(i)$
and $3x + 4y = 0 .....(ii)$
Solving these lines, we get point of intersection as $\Big(\frac{-20}{17},\frac{15}{17}\Big).$
$\therefore$ Distance of this point from the line $'5x - 2y = 0'$
$=\frac{\Big|5\times\Big(-\frac{20}{17}\Big)-2\Big(\frac{15}{17}\Big)\Big|}{\sqrt{25+4}}$
$=\frac{\Big|\frac{-100}{17}-\frac{30}{17}\Big|}{\sqrt{25}}$
$=\frac{130}{17\sqrt{29}}$
View full question & answer→
MCQ 1171 Mark
If line joining $(1, 2)$ and $(5, 7)$ is parallel to line joining $(3, 4)$ and $(11, x):$
  • A
    $10$
  • B
    $11$
  • C
    $12$
  • ✓
    $14$
Answer
Correct option: D.
$14$
We know, slope of line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{(\text{y}_{2}- \text{y}_{1})}{(\text{x}_{2}-\text{x}_{1})}$
Lines are parallel means slope is equal
$\Rightarrow\frac{(\text{x}-4)}{(11-3)} = \frac{(7-2)}{(5-1)}$
$\Rightarrow \text{x}-4 = \frac{5\times8}{4} = 10$
$\Rightarrow x = 14$
View full question & answer→
MCQ 1181 Mark
The distance between $(6, 5)$ and $(-3, 4)$ is:
  • ✓
    $\sqrt{82}$
  • B
    $\sqrt{83}$
  • C
    $\sqrt{84}$
  • D
    None
Answer
Correct option: A.
$\sqrt{82}$
The given points are $(6, 5)$ and $(-3, 4)$
The distance is given as $=\sqrt {(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt {(6+3)^2+(4-5)^2}$
$=\sqrt {9^2+1^2}$
$=\sqrt {81+1}$
$=\sqrt {82}$
View full question & answer→
MCQ 1191 Mark
Choose the correct answer. The coordinates of the foot of perpendiculars from the point $(2, 3)$ on the line $y = 3x + 4$ is given by
  • A
    $\frac{37}{10},\frac{-1}{10}$
  • ✓
    $\frac{-1}{10},\frac{37}{10}$
  • C
    $\frac{10}{37},-10$
  • D
    $\frac{2}{3},-\frac{1}{3}$
Answer
Correct option: B.
$\frac{-1}{10},\frac{37}{10}$
Let the foot of perpendicular from the point $P(2, 3)$ on the line $3x - y + 4 = 0$ be $M(h, k).$

$M(h, k)$ lies on the given line,
$\therefore 3h - k + 4 = 0 .....(i)$
Also, slopw of the given line is $3.$
$\therefore$ Slope of $\text{PM}=-\frac{1}{3}=\frac{\text{k}-3}{\text{h}-2}$ or $h + 3k - 11 = 0 .....(ii)$
Solving $(1)$ and $(ii),$ we get $(\text{h},\text{k})\equiv\Big(-\frac{1}{10},\frac{37}{10}\Big)$
View full question & answer→
MCQ 1201 Mark
If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is $(3, 2)$ then the equation of the line will be:
  • ✓
    $2x + 3y = 12$
  • B
    $3x + 2y = 12$
  • C
    $4x - 3y = 6$
  • D
    $5x - 2y = 10$
Answer
Correct option: A.
$2x + 3y = 12$
View full question & answer→
MCQ 1211 Mark
The tangent of angle between the lines whose intercepts on the axes are $a, -b$ and $b, -a$ respectively, is:
  • A
    $\frac{\text{a}^2-\text{b}^2}{\text{ab}}$
  • B
    $\frac{\text{b}^2-\text{a}^2}{\text{ab}}$
  • ✓
    $\frac{\text{b}^2-\text{a}^2}{\text{2ab}}$
  • D
    None of these
Answer
Correct option: C.
$\frac{\text{b}^2-\text{a}^2}{\text{2ab}}$
View full question & answer→
MCQ 1221 Mark
Line through the points $(-2, 6)$ and $(4, 8)$ is perpendicular to the line through the points $(8, 12)$ and $(x, 24)$. Find the value of $x:$
  • A
    $2$
  • B
    $3$
  • ✓
    $4$
  • D
    $5$
Answer
Correct option: C.
$4$
View full question & answer→
MCQ 1231 Mark
Find the distance between the following pair of points. $(5, 7)$ and the origin:
  • ✓
    $\sqrt{74}$
  • B
    $\sqrt{64}$
  • C
    $\sqrt{34}$
  • D
    None of these
Answer
Correct option: A.
$\sqrt{74}$
Take points $A (5, 7)$ and $B (0, 0)$
To find the distance between two point $A (x_1, y_1)$ and $B(x_2, y_2)$, distance formula is used which is given by:
$\text{AB} =\sqrt{ (\text{x}_{1}-\text{x}_{2})^2+(\text{y}_{1}-\text{y}_{2})^2}$
so, $\text{AB} =\sqrt{ (5-0)^{2}+(7-0)^{2}}$
$= \sqrt{(25+49)}$
$= \sqrt{(74)}$
View full question & answer→
MCQ 1241 Mark
If $p_1$ and $p_2$ are the lengths of the perpendiculars from the origin upon the lines $\text{x} \sec \theta + \text{y}\ \text{cosec}\ \theta = \text{a}$ and $\text{x} \cos \theta - \text{y} \sin \theta = \text{a} \cos 2 \theta$ respectively, then:
  • ✓
    $4\text{p}_1^2 + \text{p}_2^2 = \text{a}^2$
  • B
    $\text{p}_1^2 + 4\text{p}_2^2 = \text{a}^2$
  • C
    $\text{p}_1^2 + 4\text{p}_2^2 = \text{a}^2$
  • D
    None of these.
Answer
Correct option: A.
$4\text{p}_1^2 + \text{p}_2^2 = \text{a}^2$
The given lines are
$\text{x} \sec \theta + \text{y} \text{cosec}\theta = \text{a} \ ...(1)$
$\text{x} \cos \theta - \text{y} \sin \theta = \text{a} \cos 2 \theta \ ...(2)$
$p_1$ and $p_2$ are the perpendiculars from the origin upon the lines $(1)$ and $(2)$, respectively.
$\text{p}_1=\Big|\frac{-\text{a}}{\sqrt{\sec^2\theta+\text{cosec}^2}\theta}\Big|$ and $\text{p}_2=\Big|\frac{-\text{a}\cos2\theta}{\sqrt{\cos^2\theta+\sin^2}\theta}\Big|$
$\Rightarrow\text{p}_1=\Big|\frac{-\text{a}\sin\theta\cos\theta}{\sqrt{\sin^2\theta+\cos^2}\theta}\Big|$ and $\text{p}_2=\big|-\text{a}\cos2\theta\big|$
$\Rightarrow\text{p}_1=\frac{1}{2}\big|-\text{a}\times2\sin\theta\cos\theta\big|$ and $\text{p}_2=\big|-\text{a}\cos2\theta\big|$
$\Rightarrow\text{p}_1=\frac{1}{2}\big|-\text{a}\times\sin2\theta\big|$ and $\text{p}_2=\big|-\text{a}\cos2\theta\big|$
$\Rightarrow4\text{p}_1^2+\text{p}_2^2=\text{a}^2(\sin^22\theta+\cos^22\theta)$
$=\text{a}^2$
View full question & answer→
MCQ 1251 Mark
If slope of a line is positive then its inclination is:
  • A
    Right angle
  • ✓
    Acute angle
  • C
    Obtuse angle
  • D
    Zero
Answer
Correct option: B.
Acute angle
If inclination is $\alpha$ slope is given by $\tan\alpha$ Given that slope of line is positive which means $\tan\alpha$ is positive.
We know, $\tan\alpha$ is positive in $1^{st}$ quadrant
i.e. $\alpha$ should be acute angle
View full question & answer→
MCQ 1261 Mark
If equation of line is $x + y = 2$ then find the perpendicular distance of line from origin:
  • ✓
    $\sqrt{2}$
  • B
    $\sqrt{3}$
  • C
    $\frac{1}{\sqrt2}$
  • D
    $\frac{1}{\sqrt3}$
Answer
Correct option: A.
$\sqrt{2}$
Given, equation is $x + y = 2.$ Reducing the above equation to normal form
$\frac{(\text{x + y})}{\sqrt2} =\sqrt{2}$
$\text{x} \cos⁡45^\circ + \text{y} \sin 45^\circ = \sqrt{2}$
Perpendicular distance from origin is $\sqrt{2}$
View full question & answer→
MCQ 1271 Mark
Two lines are perpendicular if the product of their slopes is:
  • A
    $0$
  • B
    $1$
  • ✓
    $-1$
  • D
    None of these
Answer
Correct option: C.
$-1$
View full question & answer→
MCQ 1281 Mark
The figure formed by the lines $\text{ax} \pm \text{by} \pm \text{c} = 0$ is:
  • A
    a rectangle.
  • B
    a square.
  • ✓
    a rhombus.
  • D
    None of these.
Answer
Correct option: C.
a rhombus.
The given lines can be written separately in the following manner:
$ax + by + c = 0 ...(1)$
$ax + by - c = 0 ...(2)$
$ax - by - c = 0 ...(3)$
$ax - by - c = 0 ...(4)$
Graph of the given lines is given below:

Clearly, $\text{AB}=\text{BC}=\text{CD}=\text{DA}=\sqrt{\frac{\text{a}^2}{\text{c}^2}+\frac{\text{b}^2}{\text{c}^2}}=\frac{\sqrt{\text{a}^2+\text{b}^2}}{|\text{c}|}$
Thus, the region formed by the given lines is $\text{ABCD,}$ which is a rhombus.
View full question & answer→
MCQ 1291 Mark
The line segment joining the points $(1, 2)$ and $(-2, 1)$ is divided by the line $3x + 4y = 7$ in the ratio:
  • A
    $3 : 4$
  • B
    $4 : 3$
  • C
    $9 : 4$
  • ✓
    $4 : 9$
Answer
Correct option: D.
$4 : 9$
Let the line segment joining the points $(1, 2)$ and $(−2, 1)$ be divided by the line $3x + 4y = 7$ in the ratio $m:n.$
Then, the coordinates of this point will be $\Big(\frac{-2\text{m}+\text{n}}{\text{m}+\text{n}},\frac{\text{m}+2\text{n}}{\text{m}+\text{n}}\Big)$ that lie on the line.
$3\text{x}+4\text{y}=7$
$3\times\frac{-2\text{m}+\text{n}}{\text{m}+\text{n}}+4\times\frac{\text{m}+2\text{n}}{\text{m}+\text{n}}=7$
$\Rightarrow-2\text{m}+11\text{n}=7\text{m}+7\text{n}$
$\Rightarrow-9\text{m}=-4\text{n}$
$\Rightarrow\text{m}:\text{n}=4:9$
View full question & answer→
MCQ 1301 Mark
Equation of horizontal line above $x-$axis at $5$ units from $x-$axis is:
  • A
    $x = 5$
  • B
    $x = -5$
  • ✓
    $y = 5$
  • D
    $y = -5$
Answer
Correct option: C.
$y = 5$
Equation of $x-$axis is $y = 0.$ Horizontal line is parallel to $x-$axis and above it by $5$ units.
so, equation of line is $y = 5$
View full question & answer→
MCQ 1311 Mark
The line $x + y = 4$ divides the line joining the points $(-1, 1)$ and $(5, 7)$ in the ratio:
  • A
    $2 : 1$
  • ✓
    $1 : 2$ internally
  • C
    $1 : 2$ externally
  • D
    None of these
Answer
Correct option: B.
$1 : 2$ internally
View full question & answer→
MCQ 1321 Mark
The perpendicular distance of a line $4x + 3y + 5 = 0$ from the point $(-1, 2)$ is:
  • A
    $5$
  • B
    $4$
  • ✓
    $2$
  • D
    $1$
Answer
Correct option: C.
$2$
View full question & answer→
MCQ 1331 Mark
A line passes through the point $(2, 2)$ and is perpendicular to the line $3x + y = 3.$ Its $y-$intercept is:
  • A
    $\frac{1}{3}$
  • B
    $\frac{2}{3}$
  • C
    1
  • ✓
    $\frac{4}{3}$
Answer
Correct option: D.
$\frac{4}{3}$
The equation of the line perpendicular to $3x + y = 3$ is given below:
$\text{x}-3\text{y}+\lambda=0$
This line passes through $(2, 2).$
$2-6+\lambda=0$
$\Rightarrow\lambda=4$
So, the equation of the line will be
$x - 3y + 4 = 0$
$\Rightarrow\text{y}=\frac{1}{3}\text{x}+\frac{4}{3}$
Hence, the $y-$intercept is $\frac{4}{3}.$
View full question & answer→
MCQ 1341 Mark
If slope of a line is $4$ and $y-$intercept made by the line is $2$ then the equation of line will be:
  • A
    $y = 4x - 2$
  • ✓
    $y = 4x + 2$
  • C
    $y = 2x + 4$
  • D
    $y = 2x - 4$
Answer
Correct option: B.
$y = 4x + 2$
Let general equation of line be $y = m \times x + c.$
Given, $m = 4$ and $c = 2$
View full question & answer→
MCQ 1351 Mark
Which of the following points are $10$ units from the origin:
  • A
    $(6, 4)$
  • ✓
    $(-6, 8)$
  • C
    $(6, -8)$
  • D
    $(-6, -8)$
Answer
Correct option: B.
$(-6, 8)$
$\text{A} =\sqrt{ (6-0)^{2}+(4-0)^{2}} = \sqrt{52}$
$\text{B} =\sqrt{ (6-0)^{2}+(8-0)^{2}} =10$
$\text{C} =\sqrt{ (6-0)^{2}+(-8-0)^{2}} =10$
$\text{D} =\sqrt{ (-6-0)^{2}+(-8-0)^{2}} =10$
View full question & answer→
MCQ 1361 Mark
If equation of a line is $3x + 2y - 6 = 0$ then $x-$intercept is and $y-$intercept is:
  • A
    $3, 2$
  • ✓
    $2, 3$
  • C
    $2, 6$
  • D
    $3, 6$
Answer
Correct option: B.
$2, 3$
Reducing the above equation to intercept form $\frac{\text{x}}{\text{a}} +\frac{\text{y}}{\text{b}} =1,$
we get $\frac{\text{x}}{2} +\frac{\text{y}}{3} =1$
$a = 2$ which is $x-$intercept and $b = 3$ which is $y-$intercept.
View full question & answer→
MCQ 1371 Mark
What is the slope of a line which is parallel to $y-$axis?
  • A
    $-1.$
  • B
    $0.$
  • C
    $1.$
  • ✓
    Not defined.
Answer
Correct option: D.
Not defined.
If a line is parallel to $y-$axis then angle formed by it with $x-$axis is zero.
So, its inclination is $90^\circ .$
$\text{slope} = \tan 90^\circ$ which is not defined.
View full question & answer→
MCQ 1381 Mark
If line joining $(1, 2)$ and $(7, 6)$ is perpendicular to line joining $(3, 4)$ and $(11, x):$
  • A
    $12$
  • B
    $16$
  • ✓
    $-16$
  • D
    $-12$
Answer
Correct option: C.
$-16$
We know, slope of line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{(\text{y}_{2}-\text{y}_{1})}{(\text{x}_{2}-\text{x}_{1})}$
Lines are perpendicular means $m_1\times m_2= -1$
$\Rightarrow\big(\frac{\text{x}-4}{11-3}\big), \big(\frac{6-2}{7-1}\big) = -1$
$\Rightarrow (x - 4) (4) = (-1) (8) (6)$
$\Rightarrow x - 4 = -12$
$\Rightarrow x = -16.$
View full question & answer→
MCQ 1391 Mark
Equation of vertical line to the left of $y-$axis at $5$ units from $y-$axis is:
  • A
    $x = 5$
  • ✓
    $x = -5$
  • C
    $y = 5$
  • D
    $y = -5$
Answer
Correct option: B.
$x = -5$
Equation of $y-$axis is $x = 0.$ Vertical line is parallel to $y-$axis and to the left by $5$ units
so, equation of line is $x = -5$
View full question & answer→
MCQ 1401 Mark
What is the distance of $(5, 12)$ from the origin?
  • A
    $5$ units
  • B
    $8$ units
  • C
    $12$ units
  • ✓
    $13$ units
Answer
Correct option: D.
$13$ units
Let the points be $A (0, 0)$ and $B (5, 12).$
$A (0, 0) = (x_1, y_1)$
$B (5, 12) = (x_2, y_2)$
The distance between two points, $\text{AB} = [(\text{x}_{2}-\text{x}_{1})^2+(\text{y}_{2}-\text{y}_{1})^2]$
$\text{AB} = [(5-0)^{2}+(12-0)^2]$
$\text{AB}=\sqrt{(25+144)}$
$\text{AB}=\sqrt{(169)}$
$\text{AB}=13$
The distance of $(5, 12)$ from the origin is $13$ units.
View full question & answer→
MCQ 1411 Mark
If $p$ be the length of the perpendicular from the origin on the straight line $x + 2by = 2p,$ then what is the value of $b:$
  • A
    $\frac{1}{\text{p}}$
  • B
    $\text{p}$
  • C
    $\frac{1}{2}$
  • ✓
    $\frac{3}{2}$
Answer
Correct option: D.
$\frac{3}{2}$
View full question & answer→
MCQ 1421 Mark
If the lines $ax + 12y + 1 = 0, bx + 13y + 1 = 0$ and $cx + 14y + 1 = 0$ are concurrent, then $a, b, c$ are in:
  • A
    $H.P.$
  • B
    $G.P.$
  • ✓
    $A.P.$
  • D
    None of these.
Answer
Correct option: C.
$A.P.$
The given lines are
$ax + 12y + 1 = 0 ...(1)$
$bx + 13y + 1 = 0 ...(2)$
$cx + 14y + 1 = 0 ...(3)$
It is given that $(1), (2)$ and $(3)$ are concurrent.
$\begin{vmatrix} \text{a}&12&1\\\text{b}&13&1\\\text{c}&14&1\end{vmatrix}=0$
$\Rightarrow\text{a}(13-14)-12(\text{b}-\text{c})+14\text{b}-13\text{c}=0$
$\Rightarrow-\text{a}-12\text{b}+12\text{c}+14\text{b}-13\text{c}=0$
$\Rightarrow-\text{a}+2\text{b}-\text{c}=0$
$\Rightarrow2\text{b}=\text{a}+\text{c}$
Hence, $a, b$ and $c$ are in $AP.$
View full question & answer→
MCQ 1431 Mark
The relation between $a, b, a’$ and $b’$ such that the two lines $ax + by = c$ and $a’x + b’y = c’$ are perpendicular is:
  • A
    $aa¢ – bb¢ = 0$
  • ✓
    $aa¢ + bb¢ = 0$
  • C
    $ab + a¢b¢ = 0$
  • D
    $ab - a¢b¢ = 0$
Answer
Correct option: B.
$aa¢ + bb¢ = 0$
View full question & answer→
MCQ 1441 Mark
Choose the correct answer. The equations of the lines which pass through the point $(3, -2)$ and are inclined at $60^\circ$ to the line $\sqrt{3} \text{x} + \text{y} = 1$ is:
  • ✓
    $\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$
  • B
    $\sqrt{3}\text{x}+\text{y}+\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0$
  • C
    $\text{x}+\sqrt{3}\text{y}-\sqrt{3}=0,\text{x}-\sqrt{3}\text{y}-\sqrt{3}=0$
  • D
    None of these.
Answer
Correct option: A.
$\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$
Slope of the given line $\sqrt{3}\text{x}+\text{y}=1$ is, $\text{m}=-\sqrt{3}.$
Let the slope of the required line which makes an angle of $60^\circ$ with the above line is $m.$
$\therefore \tan 60^\circ=\bigg|\frac{-\sqrt{3}-\text{m}}{1-\sqrt{3}\text{m}}\bigg|$
$\Rightarrow \bigg|\frac{-\sqrt{3}-\text{m}}{1-\sqrt{3}\text{m}}\bigg|=\sqrt{3}$
$\Rightarrow -\sqrt{3}-\text{m}=\sqrt{3}-3\text{m}$ or $-\sqrt{3}-\text{m}=-\sqrt{3}+3\text{m}$
$\Rightarrow \text{m}=\sqrt{3}$ or $m = 0$
Line is passing throught the point $(3, -2).$
Thus, the equation of the required line is : $\text{y}+2=\sqrt{3}(\text{x}-3)$ or $y + 2 = 0$
$\Rightarrow \sqrt{3}\text{x}-\text{y}-2-3\sqrt{3}=0$ and $y + 2 = 0$
View full question & answer→
MCQ 1451 Mark
The lines $x + 2y - 5 = 0, 2x - 3y + 4 = 0, 6x + 4y - 13 = 0:$
  • A
    Are concurrent
  • ✓
    Form a right angled triangle
  • C
    Form an isosceles triangle
  • D
    Form an equilateral triangle
Answer
Correct option: B.
Form a right angled triangle
View full question & answer→
MCQ 1461 Mark
The vertices of a triangle are $(6, 0), (0, 6)$ and $(6, 6)$. The distance between its circumcentre and centroid is:
  • A
    $2\sqrt{2}$
  • B
    $2$
  • ✓
    $\sqrt{2}$
  • D
    $1$
Answer
Correct option: C.
$\sqrt{2}$
Let $A(0, 6), B(6, 0)$ and $C(6, 6)$ be the vertices of the given triangle.

Centroid of $\triangle\text{ABC}=\Big(\frac{0+6+6}{3},\frac{6+0+6}{3}\Big)$
$=(4,4)$
Coordinates of $\text{N}=\Big(\frac{6+6}{2},\frac{6+0}{2}\Big)$
$=(6,3)$
Coordinates of $\text{P}=\Big(\frac{0+6}{2},\frac{6+6}{2}\Big)$
$=(3,6)$
Equation of $MN$ is $y = 3$
Equation of $MP$ is $x = 3$
As, we know that circumcentre of a triangle is the intersection of the perpendicular bisectors of any two sides.
Therefore, coordinates of circumcentre is $(3, 3)$
Thus, the coordinates of the circumcentre are $(3, 3)$ and the centroid of the triangle is $(4, 4).$
Let $d$ be the distance between the circumcentre and the centroid.
$\therefore\text{d}\sqrt{(4-3)^2+(4-3)^2}=\sqrt{2}$
View full question & answer→
MCQ 1471 Mark
What is the distance between $(1, 3)$ and $(5, 6)?$
  • A
    $3$ units.
  • B
    $4$ units.
  • ✓
    $5$ units.
  • D
    $25$ units.
Answer
Correct option: C.
$5$ units.
We know, distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(\text{x}_{1}-\text{x}_{2}) ^{2}+{(\text{y}_{1}-\text{y}_{2})} ^{2}}$
So, distance between $(1, 3)$ and $(5, 6)$ is
$\sqrt{{(1-5)}^{2}+(3-6)^{2}}$
$= \sqrt{{(4)}^{2}+(3)^{2}}$
$= 5\text{ units}.$
View full question & answer→
MCQ 1481 Mark
The equation of the line passing through $(1, 5)$ and perpendicular to the line $3x - 5y + 7 = 0$ is:
  • ✓
    $5x + 3y - 20 = 0$
  • B
    $3x - 5y + 7 = 0$
  • C
    $3x - 5y + 6 = 0$
  • D
    $5x + 3y + 7 = 0$
Answer
Correct option: A.
$5x + 3y - 20 = 0$
A line perpendicular to $3x - 5y + 7 = 0$ is given by
$5\text{x}+3\text{y}+\lambda=0$
This line passes through $(1, 5)$
$5+15+\lambda=0$
$\Rightarrow\lambda=-20$
Therefore, the equation of the required line is $5x + 3y - 20 = 0.$
View full question & answer→
MCQ 1491 Mark
The reflection of the point $(4, -13)$ about the line $5x + y + 6 = 0$ is:
  • ✓
    $(-1, -14)$
  • B
    $(3, 4)$
  • C
    $(0, 0)$
  • D
    $(1, 2)$
Answer
Correct option: A.
$(-1, -14)$
Let the reflection point be $A(h, k)$
Now, the mid point of line joining $(h, k)$ and $(4, -13)$ will lie on the line $5x + y + 6 = 0$
$\therefore5\Big(\frac{\text{h}+4}{2}\Big)+\frac{\text{k}-13}{2}+6=0$
$\Rightarrow5\text{h}+20+\text{k}-13+12=0$
$\Rightarrow5\text{h}+\text{k}+19=0 \ ...(1)$
Now, the slope of the line joining points $(h, k)$ and $(4, -13)$ are perpendicular to the line $5x + y + 6 = 0.$
slope of the line $= -5$
slope of line joining by points $(h, k)$ and $(4, -13)$
$\frac{\text{k}+13}{\text{h}-4}$
$\therefore\frac{\text{k}+13}{\text{h}-4}(-5)=-1$
$\Rightarrow5\text{k}-\text{h}+60=0 \ ...(2)$
Solving $(1)$ and $(2),$ we get
$h = -1$ and $k = -14$
View full question & answer→
MCQ 1501 Mark
The equation of the locus of a point equidistant from the point $A (1, 3)$ and $B (-2, 1)$ is:
  • A
    $6x - 4y = 5$
  • ✓
    $6x + 4y = 5$
  • C
    $6x + 4y = 7$
  • D
    $6x - 4y = 7$
Answer
Correct option: B.
$6x + 4y = 5$
View full question & answer→
MCQ 1511 Mark
The value of $\lambda$ for which the lines $3\text{x} + 4\text{y} = 5, 5\text{x} + 4\text{y} =4$ and $\lambda\text{x} + 4\text{y} = 6$ meet at a point is:
  • A
    $2$
  • ✓
    $1$
  • C
    $4$
  • D
    $3$
Answer
Correct option: B.
$1$
It is given that the lines $3\text{x} + 4\text{y} = 5, 5\text{x} + 4\text{y} =4$ and $\lambda\text{x} + 4\text{y} = 6$ meet at a point. In other words, the given lines are concurrent.
$\begin{vmatrix}3&4&-5\\5&4&-4\\\lambda&4&-6 \end{vmatrix}=0$
$\Rightarrow3(-24+16)-4(-30+4\lambda)-5(20-4\lambda)=0$
$\Rightarrow-24+120-16\lambda-100+20\lambda=0$
$\Rightarrow4\lambda=4$
$\Rightarrow\lambda=1$
View full question & answer→
MCQ 1521 Mark
Find the equation of line parallel to $y-$axis and passing through $(3, 4):$
  • ✓
    $x = 3$
  • B
    $x = 4$
  • C
    $y = 4$
  • D
    $y = 3$
Answer
Correct option: A.
$x = 3$
Let general equation of line be $y = m (x - d)$
$\Rightarrow\text{x} =\frac{ \text{y}}{\text{m + d}}$ Since line is parallel to $y-$axis
so, $\text{m}=\frac{1}{0} $ or $\frac{1}{\text{m}} =0$
$\Rightarrow x = d$
$\Rightarrow x = 3$ by substituting the point $(3, 4)$
View full question & answer→
MCQ 1531 Mark
The locus of the point of intersection of lines $\text{x} \cos \text{a} + \text{y} \sin \text{a = a}$ and $\text{x} \sin \text{a - y} \cos \text{a = b} (a$ is a variable$):$
  • A
    $ 2\left(x^2+y^2\right)=a^2+b^2 $
  • B
    $ x^2-y^2=a^2-b^2 $
  • ✓
    $ x^2+y^2=a^2+b^2 $
  • D
    None of these
Answer
Correct option: C.
$ x^2+y^2=a^2+b^2 $
View full question & answer→
MCQ 1541 Mark
The centroid of a triangle is $(2, 7)$ and two of its vertices are $(4, 8)$ and $(-2, 6).$ The third vertex is:
  • A
    $(0, 0)$
  • ✓
    $(4, 7)$
  • C
    $(7, 4)$
  • D
    $(7, 7)$
Answer
Correct option: B.
$(4, 7)$
Let $A(4, 8)$ and $B(-2, 6)$ be the given vertex.
Let $C(h, k)$ be the third vertex.
The centroid of $\triangle\text{ABC}$ is $\Big(\frac{4-2+\text{h}}{3},\frac{8+6+\text{k}}{3}\Big)$
It is given that the centroid of triangle $\text{ABC}$ is $(2, 7).$
$\therefore\frac{4-2+\text{h}}{3}=2,\frac{8+6+\text{k}}{3}=7$
$\Rightarrow\text{h}=4,\text{h}=7$
Thus, the third vertex is $(4, 7).$
View full question & answer→
MCQ 1551 Mark
Find slope of line joining $(1, 2)$ and $(4, 11):$
  • A
    $\frac{1}{3}$
  • ✓
    $3$
  • C
    $9$
  • D
    $\frac{1}{9}$
Answer
Correct option: B.
$3$
We know, slope of line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{{\text{y}}_{2}-{\text{y}}_{1}}{{\text{x}}_{2}-{\text{x}}_{1}}$
So, slope of line joining $(1, 2)$ and $(4, 11)$ is $\frac{11-2}{4-1} = \frac{9}{3} = 3$
View full question & answer→
MCQ 1561 Mark
Slope of a line which cuts off intercepts of equal lengths on the axes is:
  • ✓
    $-1$
  • B
    $0$
  • C
    $2$
  • D
    $3$
Answer
Correct option: A.
$-1$
View full question & answer→
MCQ 1571 Mark
The distance between the orthocentre and circumcentre of the triangle with vertices $(1, 2), (2, 1)$ and $\Big(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\Big)$ is:
  • ✓
    $0$
  • B
    $\sqrt{2}$
  • C
    $3+\sqrt{3}$
  • D
    none of these.
Answer
Correct option: A.
$0$
Let $A(1, 2), B(2, 1)$ and $C\ \Big(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\Big)$ be the given points.
$\therefore \ \text{AB}=\sqrt{(2-1)^2+(1-2)^2}=\sqrt{2}$
$\text{BC}=\sqrt{\Big(\frac{3+\sqrt{3}}{2}-2\Big)^2,\Big(\frac{3+\sqrt{3}}{2}-1\Big)^2}=\sqrt{2}$
$\text{AC}=\sqrt{\Big(\frac{3+\sqrt{3}}{2}-1\Big)^2,\Big(\frac{3+\sqrt{3}}{2}-2\Big)^2}=\sqrt{2}$
Thus, $\text{ABC}$ is an equilateral triangle.
We know that the orthocentre and the circumcentre of an equilateral triangle are same.
So, the distance between the the orthocentre and the circumcentre of the triangle with vertices $(1, 2), (2, 1)$ and $\Big(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\Big)$ is $0.$
View full question & answer→
MCQ 1581 Mark
Choose the correct answer. A line passes through $(2, 2)$ and is perpendicular to the line $3x + y = 3.$ Its yintercept is:
  • A
    $\frac{1}{3}$
  • B
    $\frac{2}{3}$
  • C
    $1$
  • ✓
    $\frac{4}{3}$
Answer
Correct option: D.
$\frac{4}{3}$
Any line perpendicular to $3x + y = 3$
$\text{x}-3\text{y}=\lambda \ (\lambda=\text{constant})$
If is passes through the point $(2, 2)$ then
$2-3(2)=\lambda$
$\Rightarrow \lambda=-4$
$\therefore$ Required equation is $x - 3y = -4$
$\Rightarrow 3\text{y}=-\text{x}-4$
$\Rightarrow \text{y}=\frac{1}{3}\text{x}+\frac{4}{3}\big[\because \text{y}=\text{mx}+\text{c}\big]$
So, the $y-$intercept is $\frac{4}{3}.$
Hence, the correct option is $(d).$
View full question & answer→
MCQ 1591 Mark
If equation of line is $y = 5x + 10$ then find the value of $x-$intercept made by the line:
  • ✓
    $10$
  • B
    $\frac{1}{10}$
  • C
    $\frac{-1}{10}$
  • D
    $-10$
Answer
Correct option: A.
$10$
Given, equation is $y = 5x + 10.$
$Y-$intercept means value of $y$ when $x$ is zero.
​​​​​​​$y = 0 + 10$
$\Rightarrow y = 10$
View full question & answer→
MCQ 1601 Mark
The distance between the points $(a, b)$ and $(-1, -b)$ is:
  • A
    $0$
  • B
    $1$
  • C
    $\sqrt{\text{ab}}$
  • ✓
    None of the above
Answer
Correct option: D.
None of the above
Distance between two points $({\text{x}_{1, }}{\text{y}_{1}})$ and $({\text{x}_{2, }}{\text{y}_{2}})$ can be calculated using the formula
$\sqrt{{(\text{x}_{2}-\text{x}_{1})}^2+{(\text{y}_{2}-\text{y}_{1})}^2}$
Distance between the points $(a, b)$ and $(-1, -b)$
$(-1-\text{a})^2+(\text{b - b})^2 = \sqrt{1+\text{a}^2+2\text{b}+4\text{b}^2}$
View full question & answer→
MCQ 1611 Mark
The coordinates of the foot of the perpendicular from the point $(2, 3)$ on the line $x + y - 11 = 0$ are:
  • A
    $(-6, 5)$
  • ✓
    $(5, 6)$
  • C
    $(-5, 6)$
  • D
    $(6, 5)$
Answer
Correct option: B.
$(5, 6)$
Let the coordinates of the foot of the perpendicular from the point $(2, 3)$ on the line $x + y - 11 = 0$ be $(x, y)$
Now, the slope of the line $x + y - 11 = 0$ is $-1$
So, the slope of the perpendicular $= 1$
The equation of the perpendicular is given by
$y - 3 = 1(x - 2)$
$\Rightarrow x - y + 1 = 0$
Solving $x + y - 11 = 0$ and $x - y + 1 = 0,$ we get
$x = 5$ and $y = 6$
View full question & answer→
MCQ 1621 Mark
Choose the correct answer. The equation of the straight line passing through the point $(3, 2)$ and perpendicular to the line $y = x$ is:
  • A
    $x - y = 5$
  • ✓
    $x + y = 5$
  • C
    $x + y = 1$
  • D
    $x - y = 1$
Answer
Correct option: B.
$x + y = 5$
Slope of the given line $y = x$ is $1.$
Thus, slope of line perpendicular to $y = x$ is $-1.$
Line passes through the point $(3, 2).$
So, equation of the required line is:
$y - 2 = -1(x - 3)$
$\Rightarrow x + y = 5$
View full question & answer→
MCQ 1631 Mark
Two lines are given $(x - 2y)^2+ k (x - 2y) = 0$. The value of $k,$ so that the distance between them is $3$, is:
  • A
    $\text{k} = 0$
  • ✓
    $\text{k} = \underline{+} 3 \sqrt{5}$
  • C
    $\text{k}=\underline+\sqrt{5}$
  • D
    $\text{k} = 3$
Answer
Correct option: B.
$\text{k} = \underline{+} 3 \sqrt{5}$
View full question & answer→
MCQ 1641 Mark
What is the slope of a line which is parallel to $x-$axis?
  • A
    $-1.$
  • ✓
    $0.$
  • C
    $1.$
  • D
    Not defined.
Answer
Correct option: B.
$0.$
If a line is parallel to $x-$axis then angle formed by it with $x-$axis is zero.
So, its inclination is zero.
$\text{slope}= \tan 0^\circ= 0.$
View full question & answer→
MCQ 1651 Mark
If slope of a line is $4$ and $x-$intercept made by the line is $2$ then the equation of line will be:
  • ✓
    $y = 4x - 8$
  • B
    $y = 4x + 8$
  • C
    $y = 2x + 4$
  • D
    $y = 2x - 4$
Answer
Correct option: A.
$y = 4x - 8$
Let general equation of line be $y = m \times x + c.$
Given $m = 4$ and value of $x$ when $y = 0$ is $2.$
$C = -m \times 2 = -4 \times 2 = -8.$
$\Rightarrow y = 4x - 8.$
View full question & answer→
MCQ 1661 Mark
The distance between $A (-6, 7)$ and $B (-1, -5)$ is:
  • A
    $12$
  • ✓
    $13$
  • C
    $7$
  • D
    $\sqrt{37}$
Answer
Correct option: B.
$13$
Distance between $A$ and $B$ is given by
$\mid\text{A-B}\mid\mid\text{AB}\mid^{2}$
$= (-6+1)^{2}+(5+7)^{2}$
$= 25+144$
$=169$
Then $\mid\text{A - B}\mid = 13$
View full question & answer→
MCQ 1671 Mark
In a $\triangle\text{ABC}$ if $A$ is the point $(1, 2)$ and equations of the median through $B$ and $C$ are respectively $x + y = 5$ and $x = 4,$ then $B$ is:
  • A
    $(1, 4)$
  • ✓
    $(7, -2)$
  • C
    None of these
  • D
    $(4, 1)$
Answer
Correct option: B.
$(7, -2)$
View full question & answer→
MCQ 1681 Mark
Let the perpendiculars from any point on the line $7x + 56y = 0$ upon $3x + 4y = 0$ and $5x - 12y = 0$ be $p$ and $p,$ then:
  • A
    $2p = p$
  • B
    $p = 2p$
  • ✓
    $p = p$
  • D
    None of these
Answer
Correct option: C.
$p = p$
View full question & answer→
MCQ 1691 Mark
Equation of horizontal line below $x-$axis at $5$ units from $x-$axis is:
  • A
    $x = 5$
  • B
    $x = -5$
  • C
    $y = 5$
  • ✓
    $y = -5$
Answer
Correct option: D.
$y = -5$
Equation of $x-$axis is $y = 0.$
Horizontal line is parallel to $x-$axis and below it by $5$ units
so, equation of line is $y = -5$
View full question & answer→
MCQ 1701 Mark
If the projections of a line segment on the $x, y$ and $z$ axes in $3-$dimensional space are $2, 3$ and $6$ respectively, then the length of line segment is:
  • A
    $6$
  • ✓
    $7$
  • C
    $9$
  • D
    $12$
Answer
Correct option: B.
$7$
Given, projections are $2, 3$ and $6$
$\therefore\text{AB} = ({\text{x}_{1}}-{\text{x}_{2}})^2+({\text{y}_{1}}-{\text{y}_{2}})^2+({\text{z}_{1}}-{\text{z}_{2}})^2$
$\Rightarrow\text{AB} = \sqrt{4+9+36} = \sqrt{49} = 7$
View full question & answer→
Generate Paper FreeAll Straight Lines topics