MCQ 11 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
View full question & answer→Correct option: B.
$(4, 1)$
$(4, 1)$
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35 questions · auto-graded multiple-choice test.
MCQ 21 Mark
The point which divides the join of $(1, 2)$ and $(3, 4)$ externally in the ratio $1 : 1$ :
- ALies in the $III$ quadrant.
- BLies in the $II$ quadrant.
- CLies in the $I$ quadrant.
- ✓Cannot be found.
Answer
View full question & 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 $1 : 1$
$\Big(\frac{1\times3-1\times1}{1-1},\frac{1\times4-1\times2}{1-1}\Big)$
which is not defined.
Therefore, it is not possible to externally divide the line joining two points in the ratio $1 : 1$
MCQ 31 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$(1, 2)$
- Dnone of these.
Answer
View full question & 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 $y - 1 = 0.$
$\Rightarrow x = 1, y = 1$
$\therefore \ \frac{\text{a}+\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{2\text{a}+0+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{0+2\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}=0$
$\Rightarrow3\text{a}+3\text{b}+3\text{c}=0$
$\Rightarrow\text{a}+\text{b}+\text{c}=0$
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 $y - 1 = 0.$
$\Rightarrow x = 1, y = 1$
MCQ 41 Mark
The acute angle between the medians drawn from the acute angles of $a$ right angled isosceles triangle is:
- A$\cos^{-1}\big(\frac{2}{3}\big)$
- ✓$\cos^{-1}\big(\frac{3}{4}\big)$
- C$\cos^{-1}\big(\frac{4}{5}\big)$
- D$\cos^{-1}\big(\frac{5}{6}\big)$
Answer
View full question & answer→Correct option: B.
$\cos^{-1}\big(\frac{3}{4}\big)$
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 $\cos^{-1}\Big(\frac{4}{5}\Big).$
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 $\cos^{-1}\Big(\frac{4}{5}\Big).$
MCQ 51 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
View full question & 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.
$\frac{0-\text{b}}{\frac{\text{a}}{2}-0}\times\frac{-\frac{\text{b}}{2}}{-\frac{\text{a}}{2}}=-1$
$\Rightarrow\text{a}=\pm\sqrt{2}\text{b}$
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.
$\frac{0-\text{b}}{\frac{\text{a}}{2}-0}\times\frac{-\frac{\text{b}}{2}}{-\frac{\text{a}}{2}}=-1$
$\Rightarrow\text{a}=\pm\sqrt{2}\text{b}$
MCQ 61 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
View full question & 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 $m : n.$
$\therefore\Big(\frac{\text{m}-3\text{n}}{\text{m}+\text{n}},\frac{-2\text{m}-4\text{n}}{\text{m}+\text{n}}\Big)=(0,\text{t})$
$\Rightarrow0=\frac{\text{m}-3\text{n}}{\text{m}+\text{n}}$
$\Rightarrow\text{m}:\text{n}=3:1$
$\therefore\Big(\frac{\text{m}-3\text{n}}{\text{m}+\text{n}},\frac{-2\text{m}-4\text{n}}{\text{m}+\text{n}}\Big)=(0,\text{t})$
$\Rightarrow0=\frac{\text{m}-3\text{n}}{\text{m}+\text{n}}$
$\Rightarrow\text{m}:\text{n}=3:1$
MCQ 71 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$2$
- DInfinite.
Answer
View full question & 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.
$\lambda\text{x} + 3\text{y} + 1 = 0 \ ...(\text{ii})$
$4\text{x} - \lambda\text{y} + 2 = 0 \ ...(\text{iii})$
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.
MCQ 81 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
View full question & 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:
$\frac{\text{x}}{10}+\frac{\text{y}}{4}=1$
$\Rightarrow2\text{x}+5\text{y}=20$
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:
$\frac{\text{x}}{10}+\frac{\text{y}}{4}=1$
$\Rightarrow2\text{x}+5\text{y}=20$
MCQ 91 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
View full question & 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, $\text{d}=\Bigg|\frac{-7-\frac{14}{3}}{\sqrt{5^2+3^2}}\Bigg|$
$\Rightarrow\text{d}=\frac{35}{3\sqrt{34}}$
$5\text{x}+3\text{y}-7=0 \ ...(1)$
$5\text{x}+3\text{y}+\frac{14}{3}=0 \ ...(2)$
Let $d$ be the distance between the lines $5x + 3y - 7 = 0$ and $15x + 9y + 14 = 0$
Then, $\text{d}=\Bigg|\frac{-7-\frac{14}{3}}{\sqrt{5^2+3^2}}\Bigg|$
$\Rightarrow\text{d}=\frac{35}{3\sqrt{34}}$
MCQ 101 Mark
The area of a triangle with vertices at $(-4, -1), (1, 2)$ and $(4, -3)$ is :
- ✓$17$
- B$16$
- C$15$
- DNone of these.
Answer
View full question & answer→Correct option: A.
$17$
Let $A$ be the area of the triangle formed by the points $(-4, -1), (1, 2)$ and $(4, -3).$
$\therefore\text{A}=\frac{1}{2}\big|\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\}\big|$
$\Rightarrow\text{A}=\frac{1}{2}\big|\{-4(2+3)+1(-3+1)+4(-1-2)\}\big|$
$\Rightarrow\text{A}=17$
$\therefore\text{A}=\frac{1}{2}\big|\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\}\big|$
$\Rightarrow\text{A}=\frac{1}{2}\big|\{-4(2+3)+1(-3+1)+4(-1-2)\}\big|$
$\Rightarrow\text{A}=17$
MCQ 111 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$20x + 9y + 53 = 0$
- Dnone of these.
Answer
View full question & 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 :
$\frac{\text{x}}{\frac{-32}{3}}+\frac{\text{y}}{\frac{24}{5}}=1$
$\Rightarrow\frac{-3\text{x}}{32}+\frac{5\text{y}}{24}=1$
$\Rightarrow-9\text{x}+20\text{y}=96$
$\Rightarrow9\text{x}-20\text{y}+96=0$
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 :
$\frac{\text{x}}{\frac{-32}{3}}+\frac{\text{y}}{\frac{24}{5}}=1$
$\Rightarrow\frac{-3\text{x}}{32}+\frac{5\text{y}}{24}=1$
$\Rightarrow-9\text{x}+20\text{y}=96$
$\Rightarrow9\text{x}-20\text{y}+96=0$
MCQ 121 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
View full question & 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.
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.
MCQ 131 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$\text{p}^2=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
- ✓$\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
- DNone of these.
Answer
View full question & 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,
$\Rightarrow\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
$\frac{1}{\text{a}}\text{x}+\frac{1}{\text{b}}\text{y}-1=0$
$\therefore\text{p}=\begin{vmatrix}\frac{0+0+1}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}}} \end{vmatrix}$
Squaring both sides,
$\Rightarrow\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
MCQ 141 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
View full question & 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
$\frac{3}{5}:\frac{7}{5}$
$=3:7$
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
$\frac{3}{5}:\frac{7}{5}$
$=3:7$
MCQ 151 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
View full question & 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, $\tan\theta=\frac{ 2-6}{1+3}=-1$
$\Rightarrow\theta=\frac{3\pi}{4}$
Let $\theta$ be the inclination of the straight line passing through the points $(-3, 6) $ and $(1, 2)$.
Then, $\tan\theta=\frac{ 2-6}{1+3}=-1$
$\Rightarrow\theta=\frac{3\pi}{4}$
MCQ 161 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
View full question & 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 $90^\circ .$
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 $90^\circ .$
MCQ 171 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$3$
- DNone of these
Answer
View full question & 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:
$\triangle\text{DBC}:\triangle\text{ABC}=1:2$
$\frac{2(14\text{x}-7)}{49}=\frac{1}{2}$
$\Rightarrow8\text{x}-4=7$
$\Rightarrow\text{x}=\frac{11}{8}$
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:
$\triangle\text{DBC}:\triangle\text{ABC}=1:2$
$\frac{2(14\text{x}-7)}{49}=\frac{1}{2}$
$\Rightarrow8\text{x}-4=7$
$\Rightarrow\text{x}=\frac{11}{8}$
MCQ 181 Mark
The equations of the sides $AB, BC$ and $CA$ 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$3x - y + 2 = 0$
- DNone of these.
Answer
View full question & answer→Correct option: B.
$x - 3y + 4 = 0$
The equation of the sides $AB, BC$ and $CA$ of $\triangle \text{ABC}$ are $y - x = 2, x + 2y = 1$ and $3x + y + 5 = 0,$ respectively.
Solving the equations of $AB$ and $BC,$
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:
$\text{y}-1+\frac{1}{3}(\text{x}+1)$
$\Rightarrow\text{x}-3\text{y}+4=0$
Solving the equations of $AB$ and $BC,$
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:
$\text{y}-1+\frac{1}{3}(\text{x}+1)$
$\Rightarrow\text{x}-3\text{y}+4=0$
MCQ 191 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$\Big(-2,\frac{2}{3}\Big)$
- DNone of these.
Answer
View full question & 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 $\Big(\frac{2}{3},2\Big).$
$a + b + c = 0$
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 $\Big(\frac{2}{3},2\Big).$
MCQ 201 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
View full question & answer→Correct option: C.
$3$
The lines $x + q = 0, y - 2 = 0$ and $3x + 2y + 5 = 0$ are concurrent.
$\therefore\begin{vmatrix}1&0&\text{q}\\0&1&-2\\3&2&5 \end{vmatrix}=0$
$\Rightarrow1(5+4)-0+\text{q}(0-3)=0$
$\Rightarrow3\text{q}=9$
$\Rightarrow\text{q}=3$
$\therefore\begin{vmatrix}1&0&\text{q}\\0&1&-2\\3&2&5 \end{vmatrix}=0$
$\Rightarrow1(5+4)-0+\text{q}(0-3)=0$
$\Rightarrow3\text{q}=9$
$\Rightarrow\text{q}=3$
MCQ 211 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
View full question & 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 $(c)$.
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 $(c)$.
MCQ 221 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
View full question & 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, $\text{A}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$\Rightarrow\text{A}=\frac{1}{2}[(\text{a}+3)(\text{a}+4)(\text{a}+2-\text{a}-1)+(\text{a}+2)(\text{a}+3)\text{a}+1-\text{a}-3)+(\text{a}+1)(\text{a}+2)(\text{a}+3-\text{a}-2)]$
$\Rightarrow\text{A}=\frac{1}{2}[(\text{a}+3)(\text{a}+4)-2(\text{a}+2)(\text{a}+3)+(\text{a}+1)(\text{a}+2)]$
$\Rightarrow\text{A}=\frac{1}{2}[\text{a}^2+7\text{a}+12-2\text{a}^2-10\text{a}-12+\text{a}^2+3\text{a}+2]$
$\Rightarrow\text{A}=1$
Let $A$ be the area of the triangle formed by these points.
Then, $\text{A}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$\Rightarrow\text{A}=\frac{1}{2}[(\text{a}+3)(\text{a}+4)(\text{a}+2-\text{a}-1)+(\text{a}+2)(\text{a}+3)\text{a}+1-\text{a}-3)+(\text{a}+1)(\text{a}+2)(\text{a}+3-\text{a}-2)]$
$\Rightarrow\text{A}=\frac{1}{2}[(\text{a}+3)(\text{a}+4)-2(\text{a}+2)(\text{a}+3)+(\text{a}+1)(\text{a}+2)]$
$\Rightarrow\text{A}=\frac{1}{2}[\text{a}^2+7\text{a}+12-2\text{a}^2-10\text{a}-12+\text{a}^2+3\text{a}+2]$
$\Rightarrow\text{A}=1$
MCQ 231 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$2y - 3x - 2 = 0$
- DNone of these.
Answer
View full question & 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:
$\text{y}=-\frac{3}{2}\text{x}+1,$
$\Rightarrow3\text{x}+2\text{y}-2=0$
$4x + 3y - 7 = 0 ...(1)$
$8x + 5y - 1 = 0 ...(2)$
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:
$\text{y}=-\frac{3}{2}\text{x}+1,$
$\Rightarrow3\text{x}+2\text{y}-2=0$
MCQ 241 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$\text{p}_1^2 + 4\text{p}_2^2 = \text{a}^2$
- DNone of these.
Answer
View full question & 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 $\text{p}_2=\big|-\text{a}\cos2\theta\big|$
$\Rightarrow4\text{p}_1^2+\text{p}_2^2=\text{a}^2(\sin^22\theta+\cos^22\theta)$
$=\text{a}^2$
MCQ 251 Mark
The figure formed by the lines $\text{ax} \pm \text{by} \pm \text{c} = 0 $ is :
- Aa rectangle.
- Ba square.
- ✓a rhombus.
- DNone of these.
Answer
View full question & 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.
$ax + by + c = 0 ...(1)$
$ax + by - c = 0 ...(2)$
$ax - by - c = 0 ...(3)$
$ax - by - c = 0 ...(4)$
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.
MCQ 261 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
View full question & 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.
$3\text{x}+4\text{y}=7$
$3\times\frac{-2\text{m}+\text{n}}{\text{m}+\text{n}}+4\times\frac{\text{m}+2\text{n}}{\text{m}+\text{n}}=7$
$\Rightarrow-2\text{m}+11\text{n}=7\text{m}+7\text{n}$
$\Rightarrow-9\text{m}=-4\text{n}$
$\Rightarrow\text{m}:\text{n}=4:9$
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.
$3\text{x}+4\text{y}=7$
$3\times\frac{-2\text{m}+\text{n}}{\text{m}+\text{n}}+4\times\frac{\text{m}+2\text{n}}{\text{m}+\text{n}}=7$
$\Rightarrow-2\text{m}+11\text{n}=7\text{m}+7\text{n}$
$\Rightarrow-9\text{m}=-4\text{n}$
$\Rightarrow\text{m}:\text{n}=4:9$
MCQ 271 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
View full question & 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 $\frac{4}{3}.$
$\text{x}-3\text{y}+\lambda=0$
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 $\frac{4}{3}.$
MCQ 281 Mark
If the lines $ax + 12y + 1 = 0, bx + 13y + 1 = 0$ and $ax + 14y + 1 = 0$ are concurrent, then $a, b, c$ are in :
- A$H.P.$
- B$G.P.$
- ✓$A.P.$
- DNone of these.
Answer
View full question & 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 $A.P$.
$ax + 12y + 1 = 0 ...(1)$
$bx + 13y + 1 = 0 ...(2)$
$cx + 14y + 1 = 0 ...(3)$
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 $A.P$.
MCQ 291 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
View full question & 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.
$\therefore\text{d}\sqrt{(4-3)^2+(4-3)^2}=\sqrt{2}$
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.
$\therefore\text{d}\sqrt{(4-3)^2+(4-3)^2}=\sqrt{2}$
MCQ 301 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
View full question & 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 $5x + 3y - 20 = 0.$
$5\text{x}+3\text{y}+\lambda=0$
This line passes through $(1, 5)$
$5+15+\lambda=0$
$\Rightarrow\lambda=-20$
Therefore, the equation of the required line is $5x + 3y - 20 = 0.$
MCQ 311 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
View full question & 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 $k = -14$
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 $k = -14$
MCQ 321 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
View full question & 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.
$\begin{vmatrix}3&4&-5\\5&4&-4\\\lambda&4&-6 \end{vmatrix}=0$
$\Rightarrow3(-24+16)-4(-30+4\lambda)-5(20-4\lambda)=0$
$\Rightarrow-24+120-16\lambda-100+20\lambda=0$
$\Rightarrow4\lambda=4$
$\Rightarrow\lambda=1$
In other words, the given lines are concurrent.
$\begin{vmatrix}3&4&-5\\5&4&-4\\\lambda&4&-6 \end{vmatrix}=0$
$\Rightarrow3(-24+16)-4(-30+4\lambda)-5(20-4\lambda)=0$
$\Rightarrow-24+120-16\lambda-100+20\lambda=0$
$\Rightarrow4\lambda=4$
$\Rightarrow\lambda=1$
MCQ 331 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
View full question & 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 $(4, 7)$.
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 $(4, 7)$.
MCQ 341 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 :
- A$0$
- B$\sqrt{2}$
- ✓$3+\sqrt{3}$
- Dnone of these.
Answer
View full question & answer→Correct option: C.
$3+\sqrt{3}$
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 $0.$
$\therefore \ \text{AB}=\sqrt{(2-1)^2+(1-2)^2}$
$=\sqrt{2}$
$\text{BC}=\sqrt{\Big(\frac{3+\sqrt{3}}{2}-2\Big)^2,\Big(\frac{3+\sqrt{3}}{2}-1\Big)^2}$
$=\sqrt{2}$
$\text{AC}=\sqrt{\Big(\frac{3+\sqrt{3}}{2}-1\Big)^2,\Big(\frac{3+\sqrt{3}}{2}-2\Big)^2}$
$=\sqrt{2}$
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 $0.$
MCQ 351 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
View full question & 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 $y = 6$
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 $y = 6$