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:
AnswerIt 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
$y = 3$
AnswerCorrect 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$):$
AnswerCorrect 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
$(7, 7)$
AnswerCorrect 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
$\frac{1}{9}$
AnswerWe 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:
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
AnswerLet $A(1, 2), B(2, 1)$ 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
$1$
- ✓
$\frac{4}{3}$
AnswerCorrect 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
$-10$
AnswerGiven, 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
$\sqrt{\text{ab}}$
- ✓
AnswerDistance 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
$(6, 5)$
AnswerCorrect 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
$x - y = 1$
AnswerCorrect 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:
AnswerCorrect 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?
AnswerIf 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
$y = 2x - 4$
AnswerCorrect 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
$\sqrt{37}$
AnswerDistance 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
- D
$(4, 1)$
AnswerCorrect 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
AnswerCorrect 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
$y = 5$
- ✓
$y = -5$
AnswerCorrect 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:
AnswerGiven, projections are $2, 3$ and $6$
$\therefore\text{AB} = ({\text{x}_{1}}-{\text{x}_{2}})^2+({\text{y}_{1}}-{\text{y}_{2}})^2+({\text{z}_{1}}-{\text{z}_{2}})^2$
$\Rightarrow\text{AB} = \sqrt{4+9+36} = \sqrt{49} = 7$
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