MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 11

$\theta$ is the angle between

$y = 2 x + 3 , y = x + 1$ , then the value of

$\tan \theta =$

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21 $/ 5$
0%
1 $/ 3$
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5 $/ 3$
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-2

$m$ Is the slope of a line and

$m+2=m+3$ , then the possible angle between line and

$x$ -axis is

0%
${0}^{o}$
0%
${90}^{o}$
0%
${60}^{o}$
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${45}^{o}$

Explanation

Given m is the slope of the line.

Also given,

$m+2=m+3$

This equation will get satisfied only when m=

$\infty$

$\because \infty+2=\infty+3$

$\infty=\infty$

Let

$\theta$ be the angle between the line and x-axis;

$m=tan\theta$

$m=tan\infty=\dfrac{\pi }{2}$

Option (B) is correct.

$\infty$

In the given figure (not drawn to scale), the coordinates of P, Q and R are (3, 0) (0, 5) and (0, -5) respectively. Find the area of

$\Delta PQR$ .

0%
20 sq. units
0%
40 sq. units
0%
15 sq. units
0%
25/2 sq. units

$P\left(1,\,0\right),Q\left(-1,\,0\right)$ and

$R\left(2,\,0\right)$ are three points, then the locus of the point

$S\left(x,\,y\right)$ such that

${SQ}^{2}+{SR}^{2}={2SP}^{2}$ is

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$2x+3=0$
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$2x-3=0$
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$-2x+3=0$
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$-2x-3=0$

The foot of the perpendicular drawn from the origin, on the line,

$3x+y=\lambda$ is

$P$ . If the line meets

$x-$ axis at

$A$ and

$y-$ axis at

$B$ , then the ratio

$BP:PA$ is :

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$1:3$
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$3:1$
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$1:9$
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$9:1$

A line is drawn through the point

$\left(1,2\right)$ to meet to coordinate axis at

$P$ and

$Q$ such that it forms a triangle

$OPQ$ , where

$O$ is the origin. If the area of the triangle

$OPQ$ is least, then the slope of the

$PQ$ is

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$-2$
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$-1/2$
0%
$-1/4$
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$-4$

$AB$ is a distance between two points

$A(0,a)$ and

$B(0,b)$ then

$AB=$ __

0%
$|a+b|$
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$|a-b|$
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$a^{2}+b^{2}$
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$\sqrt {a^{2}+b^{2}}$

Explanation

$\begin{array}{l} \left| { AB } \right| =\sqrt { { { \left( { 0-0 } \right) }^{ 2 } }+{ { \left( { a-b } \right) }^{ 2 } } } \\ =\sqrt { { { \left( { a-b } \right) }^{ 2 } } } =\left| { a-b } \right| \end{array}$

If a vertex of an equilateral triangle is on origin and second vertex is

$(4,0)$ , then its third vertex is

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$(2,\pm\ \sqrt{3})$
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$(3,\pm\ \sqrt{2})$
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$(2,\pm\ 2\sqrt{3})$
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$(3,\pm\ 2\sqrt{2})$

Explanation

Let

$(x,y)$ be the co-ordinate of the third vertex.

Then distance between

$(x,y)$ and

$(0,0)$ is same as the distance between

$(x,y)$ and

$(4,0)$ is same as the distance between

$(0,0)$ and

$(4.0)$ .

Then we've

$\sqrt{x^2+y^2}=\sqrt{(x-4)^2+y^2}=\sqrt{4^2+0^2}$

or,

$x^2+y^2=(x-4)^2+y^2=4^2$ [ Squaring we get]

Taking the first two equality we get,

$x^2=(x-4)^2$

or,

$x=4-x$

or,

$x=2$ .

Also

$x^2+y^2=16$

or,

$y^2=12$ [ Using

$x=2$ ]

or,

$y=\pm 2\sqrt{3}$ .

So the third vertex is

$(2,\pm2\sqrt{3})$ .

Which pair has the maximum number of common points?

Explanation

$(a)$ The above circles do not intersect each other at any point.

$\therefore$ , they do not have any point in common.

$(b)$ The above circles touch each other only at one point

$Y$ .

$\therefore$ , there is

$1$ point in common.

$(c)$ The above circles touch each other at

$1$ point

$X$ only.

$\therefore$ , the circles have

$1$ point in common.

$(d)$ These circles intersect each other at two points

$G$ and

$H$ .

$\therefore$ , the circles have two points in common. It can be observed that there can be a maximum of

$2$ points in common.

The distance between

$(3,5)$ and

$(5,3)$

0%
$2\sqrt 2$
0%
$\sqrt 2$
0%
$2$
0%
None.

Explanation

Points $(3,5),(5,3)$

The distance between them is given as $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\sqrt{(3-5)^2+(5-3)^2}\\\sqrt{4+4}=\sqrt {8}=2\sqrt 2$

If area of a triangle is

$35$ square units with vertices

$\left( {2, - 6} \right),\,\,\left( {5,\,\,4} \right)$ and

$({k},\,\,4)$ then

${k}$ is :

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$-2 \space \text{or} \space 12$
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$12$
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$-2$
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$2 \space \text{or} \space 12$

The coordinates of points on the line joining the points

$P(3,-4)$ and

$Q(-2,5)$ that is twice as far from

$P$ as from

$Q$ is

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$(\dfrac{-1}{3}, 2)$
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$(\dfrac{2}{3}, 2)$
0%
$(\dfrac{1}{3}, 4)$
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$(\dfrac{4}{3}, 2)$

Explanation

The given co-ordinates

$P=(3,-4)$ and

$Q=(-2,5)$

We have to find the co-ordinates of the point that is dividing the line joining

$P$ and

$Q$ in the ratio

$2:1$

Let the co-ordinates of the point be

$(x,y)$

If co-ordinates of

$P$ be

$(x_1,y_1)$ and c-o-ordinates of

$Q$ be

$(x_2,y_2$ and the ratio in which

$P$ and

$Q$ is dividing by a point

$(x,y)$ is

$m_1:m_2$ then,

$x=\dfrac{m_1x_2+m_2x_1}{m_1+m_2}$

$\Rightarrow$

$x=\dfrac{2\times (-2)+1\times 3}{1+2}$

$\Rightarrow$

$x=\dfrac{-4+3}{1+2}$

$\Rightarrow$

$x=\dfrac{-1}{3}$

$y=\dfrac{m_1y_2+m_2y_1}{m_1+m_2}$

$\Rightarrow$

$y=\dfrac{2(5)+1(-4)}{2+1}$

$\Rightarrow$

$y=\dfrac{10-4}{3}$

$\Rightarrow$

$y=\dfrac{6}{3}$

$\therefore$

$y=2$

$\therefore$ The co-ordinates of the point is

$\left(\dfrac{-1}{3},2\right)$

$(5,-4)$ and

$(-3,2)$ are two opposite vertices of square then its area is -

0%
$50$
0%
$75$
0%
$25$
0%
$100$

Explanation

Given

$(5,-4)$ &

$(-3,2)$ are two opposite vertices of square then.

$AC=\sqrt{(5+3)^2+(-4-2)^2}$

$AC=\sqrt{64+36}=10cm$

Now in

$\triangle ADC$

$x^2+x^2=100$

$x^2=50$

[x=side of square]

Hence area of square

$=50$ sq. unit.

2118117_1488651_ans_5f9639401c494150bcc603cafce584ca.png

The angle between the straight lines

$x - y \sqrt { 3 } = 5$ and

$x \sqrt { 3 } + y = 7$ is.

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$90 ^ { \circ }$
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$60 ^ { \circ }$
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$75 ^ { \circ }$
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$30 ^ { \circ }$

The distance of the point

$(1, 2)$ from the line

$x+y=0$ measures parallel to the line

$3x-y=2$ is

0%
$10$
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$\dfrac{3\sqrt {10}}{4}$
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$\dfrac{3\sqrt {2}}{8}$
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$5\sqrt 5$

Area of the triangle with vertices

$(-2,2), (1,5)$ and

$(6,-1)$ is

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$15$
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$\dfrac{3}{5}$
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$\dfrac{29}{2}$
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$\dfrac{33}{2}$

Explanation

Area of triangle having vertices

$(x_1,y_1), (x_2,y_2)$ and

$(x_3,y_3)$ is given by
Area

$= \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$

$=\dfrac{1}{2} [-2(5+1) + 1(-1-2)+6(2-5)]$

$=\dfrac{1}{2}[-12-3-18]$

$=-\dfrac{33}{2}$

$\therefore$ Area

$=\dfrac{33}{2}~sq.\, units$

The area of a triangle with vertices

$A(5,0),B(8,0)$ and

$C(8,4)$ in square units is

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$20$
0%
$12$
0%
$6$
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$16$

Explanation

Area of Triangle using coordinates of A, B and C
Area

$= \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$ --------(1)
Here in our case

$A(5,0) = (x_1, y_1)$

$A(8,0) = (x_2, y_2)$

$A(8,4) = (x_3, y_3)$

Substituting above values in Eq - 1, we get

Area

$= 0.5 [ 5(0-4) + 8(4-0) + 8(0-0)]$
Area

$= 0.5 [ -20 + 32]$
Area

$= 0.5 \times (12)=6$

$Area = 6$ square units.

Let

$A(6,-1)\, B(1,3)$ and

$C(x,8)$ be three points such that

$AB = BC$ . The values of

$x$ are

0%
$3,5$
0%
$-3,5$
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$3,-5$
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$4,5$

Explanation

We have,

$A(6-1), B(1,3),C(x,8)$

Also,

$AB = BC$

$\Rightarrow AB^2 = BC^2$

$\Rightarrow (1-6)^2 + (3+1)^2 = (x-1)^2 + (8-3)^2$

$\Rightarrow 25 + 16 = (x-1)^2 + 25$

$\Rightarrow (x-1)^2 = 16$

$\Rightarrow x-1 = \pm 4$

$\Rightarrow x = 1 \pm 4$

$\Rightarrow x = 5, -3$

The vertices of a

$\triangle ABC$ are

$A(3,8),B(-4,2)$ and

$C(5,-1)$ . The area of

$\triangle ABC$ is

0%
$57$ sq units
0%
$75$ sq units
0%
$28\cfrac{1}{2}$ sq units
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$37\cfrac{1}{2}$ sq units

Explanation

Here

$\left( { x }_{ 1 }=3,{ y }_{ 1 }=8 \right) ;\left( { x }_{ 2 }=-4,{ y }_{ 2 }=2 \right) ;\left( { x }_{ 3 }=5,{ y }_{ 3 }=-1 \right)$

$Ar\left( \triangle ABC \right) =\cfrac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right]$

$=\cfrac { 1 }{ 2 } \left[ 3(2+1)-4(1-8)+5(8-2) \right] =\cfrac { 1 }{ 2 } (9+36+30)=\cfrac { 75 }{ 2 } =37\cfrac { 1 }{ 2 }$

If the point

$P(x, y)$ be equidistant from the points

$A(a+b, a-b)$ and

$B(a-b, a+b)$ then

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$ax=by$
0%
$bx=ay$
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$x^2-y^2=2(ax+by)$
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$P$ can be $(a, b)$

$A$ and

$B$ are two points having co-ordinates

$(3,4)$ and

$(5,-2)$ respectively and

$P$ is a point such that

$PA=PB$ and area of triangle

$PAB=10$ square units, then the co-ordinates of

$P$ are

0%
$(7,4)$ or $(13,2)$
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$(7,2)$ or $(1,0)$
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$(2,7)$ or $(4,13)$
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$None\ of\ these$

the area of a triangle with vertices

$(-3, 0),(3,0)$ and

$(0, k)$ is

$9$ sq units. then the value of

$k$ will be

0%
$9$
0%
$3$
0%
$-9$
0%
$6$

Explanation

We know that , area of a triangle with vertices

$(a_1,y_1),(x_2,y_2)$ and

$(x_3,y_3)$ is given by

$\Delta = \frac {1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|$

$\therefore \Delta = \frac {1}{2} \left| \begin{matrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} \right|$
Expanding along

$R_1$

$9 = \frac {1}{2} [ -3 ( -k) - 0 +1 ( 3k) ]$

$\Rightarrow 18 = 3k + 3k = 6 k$

$\therefore K = \frac {18 }{6} = 3$

Fill in the blank using correct alternative.
Seg AB is parallel to Y-axis and coordinates of point A are (1,3) then co-ordinates of point B can be.......

0%
(3,1)
0%
(5,3)
0%
(3,0)
0%
(1,-3)

Explanation

Slope of y-axis will be not defined because the denominator will be 0.

Since AB is parallel to y-axis, so the slope will be same.

Let the coordinate of point B be (x, y).

Slope of

$AB =\dfrac{y-3}{x-1}$ slope of y-axis Slope of y-axis is not defined as the denominator of

$\dfrac{y-3}{x-1}$ will also be 0.

so,

$x-1=0$

$\Rightarrow x=1$

With x=1 here is only 1 option that (D).

Hence, the correct answer is (1,-3) .

Fill in the blank using correct alternative. A line makes an angle of 30° with the positive direction of X— axis. So the slope of the line is _____.

0%
$\dfrac{1}{2}$
0%
$\dfrac{\sqrt{3}}{2}$
0%
$\dfrac{1}{\sqrt{3}}$
0%
$\sqrt{3}$

Explanation

Slope of line will be

$\tan \theta=tan\,30^0 =\dfrac{1}{\sqrt{3}}$

Hence, the correct answer is

$\dfrac{1}{\sqrt{3}}$

The angle between x-axis and y-axis is

0%
$0^{\circ}$
0%
$45^{\circ}$
0%
$60^{\circ}$
0%
$90^{\circ}$

Explanation

x-axis is horizontal axis and y-axis is vertical axis.

Thus, angle between them must be

${ 90 }^{ 0 }$

Line joining the points

$(1,0)$ and

$(-2, \sqrt 3)$ makes and angle

$\theta$ with

$x-$ axis then value of

$\tan \theta$ is:

0%
$\sqrt 3$
0%
$-\sqrt 3$
0%
$\dfrac {1}{\sqrt 3}$
0%
$\dfrac {1}{-\sqrt 3}$

Explanation

Slope of line joining the points

$(1,0)$ and

$(-2,\sqrt 3)$

$m=\tan \theta$

$=\dfrac{y_2-y_1}{x_2-x_1}$

$=\dfrac {\sqrt 3 -0}{-2-1}=\dfrac {\sqrt 3}{-3}$ or

$-\dfrac {1}{\sqrt3}$

Let

$PQR$ be a right angled isosceles traingle, right angled at

$P(2,1).$ If the equation of the line

$QR$ is

$2x+y=3$ , then the equation represnting the pair of lines

$PQ$ and

$PR$ is

0%
$3x^2-3y^2+8xy+2x+10y+25=0$
0%
$3x^2-3y^2+8xy-20x-10y+25=0$
0%
$3x^2-3y^2+8xy+10x+15y+20=0$
0%
$3x^2-3y^2-8xy-10x-15y-20=0$

Explanation

Let

$m$ be the slope of

$PQ$ them

$\displaystyle\tan { { 45 }^{o}= } \left| \frac { m-\left( -2 \right) }{ 1+m\left( -2 \right) } \right|$

$\displaystyle\Rightarrow 1=\left| \frac { m+2 }{ 1-2m } \right| \Rightarrow \pm 1=\frac { m+2 }{ 1-2m }$

$\displaystyle\Rightarrow m+2=1-2m$ or

$-1+2m=m+2$

$\displaystyle m=-\frac { 1 }{ 3 }$ or

$m=3$

$PQ$ markes

$\displaystyle { 45 }^{ o }$ with

$QR$

Equation of

$PQ$ is

$\displaystyle y-1=-\frac { 1 }{ 3 } \left( x-2 \right)$

$\displaystyle\Rightarrow x+3y-5=0$

Equation of

$PR$ is

$\displaystyle y-1=3\left( x-2 \right)$

$\displaystyle \Rightarrow 3x-y-5=0$

Combined equation of

$PQ$ and

$PR$ is

$\displaystyle \left( x+3y-5 \right) \left( 3x-y-5 \right) =0$

$\displaystyle \Rightarrow 3{ x }^{ 2 }-3{ y }^{ 2 }+8xy-20x-10y+25=0$

$A(-2,4)$ ,

$B(0,0)$ and

$C(4,2)$ are the vertices of a

$\Delta ABC$ , then find the length of median through the vertex A.

0%

2
0%
4
0%
3
0%
5

Explanation

A median of a triangle is a line segment that joins the vertex of a triangle to the midpoint of the opposite side.

Mid point of two points

${ (x }_{ 1 },{ y }_{ 1 })$ and

${ (x }_{ 2 },{ y }_{ 2 })$ is calculated by the formula

$\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right)$

Using this formula,

mid point of BC

$D = \left( \frac { 0 + 4}{ 2 } ,\frac { 0 + 2 }{ 2 } \right) \quad =\quad (2,1)$

Distance between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula $\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$

Length of the median A $(-2,4)$ and D $(2,1) = \sqrt { \left( 2 + 2 \right) ^{ 2 }+\left( 1 - 4 \right) ^{ 2 } } = \sqrt { 16 + 9 } = \sqrt { 25 } = 5$

$a, b, c$ are in A.P. and the points

$A(a, 1), B(b, 2)$ and

$C(c, 3)$ are such that

$(OA)^{2}, (OB)^{2}$ and

$(OC)^{2}$ are also in A.P;

$O$ being the origin, then

0%
$\displaystyle a^{2}+c^{2}=2b^{2}-2$
0%
$\displaystyle ac=b^{2}+1$
0%
$\displaystyle (a+c)^{2}=4b^{2}$
0%
$\displaystyle a+b+c=3b$

Explanation

a,b,c are in A.P
then

$2b=a+c$
now

${OA}^{2},{OB}^{2},{OC}^{2}$ are also in A.P

$\therefore 2{OB}^{2}={OA}^{2}+{OC}^{2}$
Now using distance formula,

${OA}^{2}={a}^{2}+1$

${OB}^{2}={b}^{2}+4$

${OA}^{2}={c}^{2}+9$

$2({b}^{2}+4)={a}^{2}+1+{c}^{2}+9$
On solving, we get

$2{b}^{2}={a}^{2}+{c}^{2}+2$

${a}^{2}+{c}^{2}=2{b}^{2}-2$ ---------(A)
Again

$2{b}^{2}={a}^{2}+{c}^{2}+2$

$2{b}^{2}={(a+c)}^{2}-2ac+2$ (since

$a+c=2b$ )

$2{b}^{2}={(2b)}^{2}-2ac+2$
on solving we get

$ac={b}^{2}+1$ ---- (B)
Now

$2{b}^{2}={a}^{2}+{c}^{2}+2$

$2{b}^{2}={(a+c)}^{2}-2ac+2$

$2{b}^{2}={(a+c)}^{2}-2({b}^{2}+1)+2$ by putting the value of ac obtained in B
on solving, we get

${(a+c)}^{2}=4{b}^{2}$ --- (C)

$2b=a+c$
adding b to both sides then we get

$3b=a+b+c$

Two medians drawn from acute angles of a right angled triangle intersect at an angle

$\dfrac\pi6$ . If the length of the hypotenuse of the triangle is

$3$ units, then area of the triangle (in sq. units) is

0%
$\sqrt{3}$
0%
$3$
0%
$\sqrt{2}$
0%
$9$

Explanation

$G\equiv \left(\dfrac{b}{3},\dfrac{a}{3}\right)$

Slope of line AG

$=\dfrac{a-\dfrac{a}{3}}{0-\dfrac{b}{3}}=\dfrac{-2a}{b}$

Slope of MG

$=\dfrac{\dfrac{a}{2}-\dfrac{a}{3}}{0-\dfrac{b}{3}}=\dfrac{-a}{2b}$

$\tan 30^o=\dfrac{\dfrac{-a}{2b}-(\dfrac{-2a}{b})}{1+\dfrac{a^2}{b^2}}$

$\implies \dfrac{1}{\sqrt3}=\dfrac{3ab}{2(a^2+b^2)}$

$\implies \dfrac{1}{2} ab=\dfrac{(a^2+b^2)}{3\sqrt3}$

In a right angled triangle

$AC^2+BC^2=AB^2$

$\implies a^2+b^2=9$

$\therefore Area =\dfrac{1}{2} ab=\dfrac{9}{3\sqrt3}=\sqrt3$ sq.units

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 11 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 11 - MCQExams.com

0:0:2

Practice Class 11 Engineering Maths Quiz Questions and Answers

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