MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 8

In the figure,

$AC = 9, BC = 3$ and

$D$ is

$3$ times as far from

$A$ as from

$B$ . What is

$BD$ ?

0%
6
0%
9
0%
12
0%
15
0%
18

Explanation

Let

$BD=x$

It is given that

$AC=9$ ,

$BC=3$ and

$D$ is

$3$ times as far from

$A$ as from

$B$ , therefore

$AD=3BD$ which means:

$AD=3BD$

$\Rightarrow 9+3+x=3x$

$\Rightarrow 12+x=3x$

$\Rightarrow 3x-x=12$

$\Rightarrow 2x=12$

$\Rightarrow x=6$

Hence,

$BD=6$

If the area of the triangle shown is

$30$ square units, what is the value of

$y$ ?

0%
$5$
0%
$6$
0%
$8$
0%
$12$

Explanation

Area of triangle

$=\dfrac 12 \left| \begin {matrix} 1 & 1 & 1 \\ 0 & 0 & 5 \\ y & 0 & 0 \end {matrix}\right|$

$\Rightarrow 30=\dfrac 12 \left| 1(0-0) - 1(5y-0) +1(0-0)\right|$

$\Rightarrow 60 = |-5y|$

$\Rightarrow 60 = 5y$

$\Rightarrow y =12$

If the distance between the two points

$P(a,3)$ and

$Q(4,6)$ is

$5$ , then find

$a$ .

0%
$0$
0%
$-4$
0%
$8$
0%
$-4$ and $0$
0%
$0$ and $8$

Explanation

The distance between points

$(a,3)$ and

$(4,6)$ is

$\sqrt { { (a-4) }^{ 2 }+{ (3-6) }^{ 2 } } =\sqrt { { (a-4) }^{ 2 }+9 }$
Given that the distance is

$5$
So, we have

$\sqrt { { (a-4) }^{ 2 }+9 } =5$

$\Rightarrow {(a-4)}^{2}+9=25$

$\therefore a-4 = \sqrt{16} = |4|$
So, we get

$a = 4+|4| = 0 , 8$

In the

$xy$ -plane, the vertices of a triangle are

$(-1,3), (6,3)$ and

$(-1,-4)$ . The area of the triangle is ___ square units.

0%
$10$
0%
$17.5$
0%
$24.5$
0%
$35$

Explanation

$(x_1,y_1)$ ,

$(x_2,y_2)$ and

$(x_3,y_3)$ are the vertices of a triangle then its area is given by,

$A=\left| \dfrac { 1 }{ 2 } (x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 })) \right|$

Therefore, with the vertices

$(-1,3)$ ,

$(6,3)$ and

$(-1,-4)$ .

Area of triangle is given by,

$A=\left| \dfrac { 1 }{ 2 } (-1(3-(-4))+6(-4-3)+(-1)(3-3)) \right|$

$=\left| \dfrac { 1 }{ 2 } (-7-42) \right|$

$=\left| \dfrac { 1 }{ 2 } (-49) \right|$

$=\left| -24.5 \right|$

$=24.5$ square units.

If figure

$\Box ABCD$

ABCD is a parallelogram, what is the x-coordinate of point B?

0%
$2$
0%
$5$
0%
$6$
0%
$8$

Explanation

Looking at the figure, we can say that the

$y-$ coordinate of point

$B$ is

$5$ .

Hence, the coordinates of

$B$ are

$(x,5)$ .

Since,

$\Box ABCD$ is a paralleogram,

$AB = CD$ ...Opposite sides of parallelogram are congruent

By distance formula,

$(6-2)^2+(-3+3)^2 = (x-4)^2+(5-5)^2$

$\Rightarrow 4^2 = x^2-8x+4^2$

$\Rightarrow x^2-8x=0$

$x(x-8) = 0$

$x=0$ and

$x=8$

Now,

$x \neq 0$ since point B is in the 1st quadrant.

Hence, the

$x-$ coordinate of B is

$8$ .

In the figure, there is a regular hexagon with sides of length

$6$ . If the coordinate of

$A$ is

$(9,0)$ , what is the y-coordinate of

$B$ ?

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

Explanation

Let the vertex between A and B be C (other than 'A' lying on the x-axis).

Since it is a regular hexagon, hence the length of side AC will be

$6$ .

Therefore,

$AC=6$ . Let the coordinate of C be

$(x,0)$ .

Therefore, application of distance formula gives us

$9-x=6$ or

$x=3$ .

Now let the coordinate of B be

$(0,y)$ .

Therefore,

$BC=6$

$\Rightarrow \sqrt{3^{2}+y^{2}}=6$

$\Rightarrow 9+y^{2}=36$

$\Rightarrow y^{2}=27$

$\Rightarrow y=3\sqrt{3}$ .
Hence

$B=(0,3\sqrt{3})$ .

Let

$S$ be the set of points whose abscissas and ordinates are natural numbers. Let

$P \in S$ such that the sum of the distance of

$P$ from

$(8,0)$ and

$(0,12)$ is minimum among all elements in

$S$ . Then the number of such points

$P$ in

$S$ is

0%
$1$
0%
$3$
0%
$5$
0%
$11$

Explanation

The sum of the distances of

$P$ from

$A(8,0),B(0,12)$ is minimum if

$P$ lies on line joining

$AB$ and between them

$AP+PB=AB$

$AP'+P'B>AB$ by triangle property

So let

$P(x,y)$

The line

$AB$ is

$\dfrac { x }{ 8 } +\dfrac { y }{ 12 } =1$

$3x+2y=24$

$\Rightarrow 3x=2(12-y)$

$y$ can be

$3,6,9$

$y=3,x=6\rightarrow 1$ point

$y=6,x=4\rightarrow 1$ point

$y=9,x=2\rightarrow 1$ point

So in total three points.

Which of the following sets of points is collinear

0%
$(1, -1), (-1, 1), (0, 0)$
0%
$(1, -1), (-1, 1), (0, 1)$
0%
$(1, -1), (-1, 1), (1, 0)$
0%
$(1, -1), (-1, 1), (1, 1)$

Explanation

General equation of a line:

$y = mx + c$

Now, equation of line passing through

$( 1,-1)$ and

$(-1,1)$

Here, both points

$(1,-1)$ and

$(-1,1)$ should satisfies the line.

$(1,-1)$ gives

$-1 = m+c$ ... (i)

$(-1,1)$ gives

$1 = -m +c$ ...(ii)

Solving (i) and (ii), we get

$m = -1$ and

$c=0$

Therefore equation of line is

$y = -x$

Now,

$(0,0)$ satisfies the line but

$(0,1), (1,0)$ and

$(1,1)$ does not satisfies it.

Hence,

$(1,-1), (-1,1), (0,0)$ is a set of collinear points.

Area of the triangle formed by the points

$\left( 0,0 \right) ,\left( 2,0 \right)$ and

$\left( 0,2 \right)$ is

0%
$1$ sq. unit
0%
$2$ sq. units
0%
$4$ sq. units
0%
$8$ sq. units

Explanation

Given:

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

$B=(x_2,y_2)=(2,0)$ and

$C=(x_3,y_3)=(0,2)$

Area of triangle

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

$=\dfrac {1}{2}[0(0-2)+2(2-0)+0(0-0)]$

$=\dfrac {1}{2}[2(2)]$

$=2$ sq. units.

The vertices of

$\triangle {ABC}$ are

$A(1,8),B(-2,4), C(8,-5)$ . If

$M$ and

$N$ are the midpoints of

$AB$ and

$AC$ respectively, find the slope of

$MN$ and hence verify that

$MN$ is parallel to

$BC$ .

0%
$-\cfrac{9}{10}$
0%
$\cfrac{9}{10}$
0%
$-\cfrac{9}{5}$
0%
None of these

Explanation

Given,

Vertices of triangle

$ABC$ i.e.

$A (1, 8), B (–2, 4), C (8, –5)$

$M$ and

$N$ are

$mid – points$ of

$AB$ and

$AC$ .

Finding co–ordinates of

$M$ and

$N$ ,

We know that,

$M$ is the

$mid–point$ of

$AB$

$x_1 = 1, x_2 = –2$

$y_1 = 8, y_2 = 4$

Mid–point of

$AB$

$=\left(\dfrac{1-2}{2}, \dfrac{8+4}{2}\right)$

$=\left(\dfrac{-1}{2}, \dfrac{12}{2}\right)$

$=\left(-\dfrac{1}{2}, 6\right)$

$N$ is the

$mid–point$ of

$AC$

$x_1 = 1, x_2 = 8$

$y_1 = 8, y_2 = –5$

Mid–point of

$AC$

$=\left(\dfrac{1+8}{2}, \dfrac{8-5}{2}\right)$

$=\left(\dfrac{9}{2}, \dfrac{3}{2}\right)$

Slope of

$MN$ :

Slope of line passing through

$(x_1, y_1)$ and

$(x_2, y_2)$ is

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

So,

$m_1=\dfrac{\dfrac{3}{2}-6}{\dfrac{9}{2}+\dfrac{1}{2}}$

$m_1=\dfrac{-\dfrac{9}{2}}{5}$

$m_1=-\dfrac{9}{10}$

Verification of

$MN$ and

$BC$ are parallel:

$MN$ and

$BC$ are parallel, then their slopes must be equal.

Slope of

$BC$ :

$B (–2, 4)$ and

$C (8, –5)$

Slope of

$BC$

$m_2=\dfrac{-5-4}{8+2}$

$m_2=-\dfrac{9}{10}$

$\Rightarrow m_1=m_2$

$\therefore\$ Slope of

$MN =$ Slope of

$BC$

Hence,

$MN$ is parallel to

$BC.$

What is the perimeter of the triangle with vertices

$A (-4, 2), B (0, -1)$ and

$C (3, 3)$ ?

0%
$7 + 3 \sqrt{2}$
0%
$10 + 5 \sqrt{2}$
0%
$11 + 6 \sqrt{2}$
0%
$5 + 10 \sqrt{2}$

Explanation

Given :

$A (-4, 2), B (0, -1)$ and

$C (3, 3)$

Now distance between A and B

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

Distance between A and C

$=\sqrt { (7)^{ 2 }+(1)^{ 2 } } = 5\sqrt { 2 }$

Distance between B and C

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

So perimeter

$=AB+BC+AC=10+5\sqrt2$

Hence, option B is correct.

The vertices of a triangle ABC are A(2, 3, 1), B(-2, 2, 0) and C(0, 1, -1). Find the cosine of angle ABC.

0%
$\frac{1}{\sqrt{3}}$
0%
$\frac{1}{\sqrt{2}}$
0%
$\frac{2}{\sqrt{6}}$
0%
None of the above

Explanation

Given

$A(2, 3, 1), B(-2, 2, 0)$ and

$C(0, 1, -1)$

For

$\triangle ABC,$

$a=BC=\sqrt { { (-2-0) }^{ 2 }+{ (2-1) }^{ 2 }+{ (0+1) }^{ 2 } } \\ \therefore a=\sqrt { 6 } \\ b=AC=\sqrt { { (2-0) }^{ 2 }+{ (3-1) }^{ 2 }+{ (1+1) }^{ 2 } } \\ \therefore b=\sqrt { 12 } \\ c=AB=\sqrt { { (2+2) }^{ 2 }+{ (3-2 })^{ 2 }+{ (1) }^{ 2 } } \\ \therefore c=\sqrt { 18 } \\ \cos { \angle ABC } =\cfrac { { a }^{ 2 }+{ c }^{ 2 }-{ b }^{ 2 } }{ 2ac } \\ =\cfrac { 18+6-12 }{ 2\times 3\sqrt { 2 } \times \sqrt { 6 } } \\ =\cfrac { 1 }{ \sqrt { 3 } }$

Area of the triangle formed by the points

$(0,0),(2,0)$ and

$(0,2)$ is

0%
$1$ sq.units
0%
$2$ sq.units
0%
$4$ sq.units
0%
$8$ sq.units

Explanation

Area of triangle having vertices

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

$(x_3,y_3)$ is given by
Area

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

Therefore,

$=\dfrac{1}{2}\left|0(0-2)+2(2-0)+0(0-0)\right|$

$=\dfrac{1}{2}\left|0+4+0\right|$

$=\dfrac{1}{2}\times 4=2\ sq.\ units$

Hence, this is the answer.

Find the value of

$a$ if area of the triangle is

$17$ square units whose vertices are

$(0,0), (4,a), (6,4)$ .

0%
$a=-2$
0%
$a=-3$
0%
$a=3$
0%
none of the above

Explanation

vertices of the triangle are

$A(0,0), B(4,a), C(6,4)$ .

Then area of triangle

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

$\Rightarrow 17=\dfrac{1}{2}[0(a-4)+4(4-0)+6(0-a)$

$\Rightarrow 34=16-6a$

$\Rightarrow a=-\dfrac{18}{6}=-3$

The slope of a straight line passing through A( -2, 3) is -4/The points on the line that are 10 units away from A are

0%
(-8, 11), (4, -5)
0%
(- 7, 9), (17,-1)
0%
(7, 5) (- 1, -1)
0%
(6, 10), (3, 5)

Explanation

The equation line will be

$y-3=\frac{-4}{3}(x+2)$

$4x+3y=1$

Now, let the point

$10$ unit away is

$(h,k)$

Then,

$4h+3k=1$ ------

$(1)$

And

$\sqrt {(h+2)^2+(k-3)^2}=10$

${(h+2)^2+(k-3)^2}=10$ ------

$(2)$

Now, from

$(1)$

$h=\frac{1-3k}{4}$

Put in equation

$(2)$ , we get

$25k^2-150k-1663=0$

On solving quadric equation

$K=11$ and

$k=-5$

Put

$k=11$ in equation

$(1)$

$h=-8$

And put

$k=-5$ in equation

$(1)$

$h=4$

Hence the points are

$(-8,11),(4,-5)$

Hence, the correct answer is

$A$ .

If the equation to the locus o points equidistant from the points (-2,3),(6,-5) is

$ax+by+c=0$ where

$a> 0$ then, the ascending order of a,b,c is

0%
a,b,c
0%
c,b,a
0%
b,c,a
0%
a,c,b

Explanation

$A(-2,3);B(6,-5)$

$p(x,y)$

$PA^{2}=PB^{2}$

$(x+2)^{2}+(y-3)^{2}=(x-6)^{2}+(y+5)^{2}$

$4x+4+9-6y=-12x+36+10y+25$

$16x-16y-48=0$

$x-y-3=0$

$ax+by+c=0$

$a=1;b=-1;c=-3$
Ascending order :

$c,b,a$

The point on the line

$4x+3y=5$ , which is equidistant from

$\left( 1,2 \right)$ and

$\left( 3,4 \right)$ , is

0%
$\left( 7,-4 \right)$
0%
$\left( -10,15 \right)$
0%
$\left( \dfrac { 1 }{ 7 } ,\dfrac { 8 }{ 7 } \right)$
0%
$\left( 0,\dfrac { 5 }{ 4 } \right)$

Explanation

Let the point

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ be on the line

$4x+3y=5$ .

$\therefore 4{ x }_{ 1 }+3{ y }_{ 1 }=5$ ....(i)
Also, we have

${ \left( { x }_{ 1 }-1 \right) }^{ 2 }+{ \left( { y }_{ 1 }-2 \right) }^{ 2 }={ \left( { x }_{ 1 }-3 \right) }^{ 3 }+{ \left( { y }_{ 1 }-4 \right) }^{ 2 }$

$\Rightarrow { x }_{ 1 }^{ 2 }+1-2{ x }_{ 1 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }={ x }_{ 1 }^{ 2 }+9-6{ x }_{ 1 }+{ y }_{ 1 }^{ 2 }+16-8{ y }_{ 1 }$

$\Rightarrow 4{ x }_{ 1 }+4{ y }_{ 1 }=20$

$\Rightarrow { x }_{ 1 }+{ y }_{ 1 }=5$ ....(ii)
From equations (i) and (ii), we get

${ y }_{ 1 }=15$ and

${ x }_{ 1 }=-10$

$A=(4, 2)$ and

$B=(2, 4)$ are two given points and a point

$P$ on the line

$3x + 2y + 10 = 0$ is given then. which of the following is true.

0%
$(PA + PB)$ is minimum when $P (\dfrac {-14}{5}, \dfrac {-4}{5})$
0%
$(PA + PB)$ is maximum when $P (\dfrac {-14}{5}, \dfrac {-4}{5})$
0%
$(PA - PB)$ is maximum when $(-22, 28)$
0%
$(PA - PB)$ is minimum when $P (-22, 28)$

If the slop of one of the lines represented by $ax^2-6xy+y^2=0$ is the square of the other,then the value of a is

0%
$-27$ or $8$
0%
$-3$ or $2$
0%
$-64$ or $27$
0%
$-4$ or $3$

Explanation

$m+m^2=6$

$m^3=a$

$m^3+m^6+3.m^3.6=6^3$

$a^2+19a-216=0$

$a=-27$ or

$8$

Find the area (in square units) of the triangle whose vertices are

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

$(-a, c).$

0%
$2ac$
0%
$2bc$
0%
$b(a+c)$
0%
$c(a-h)$

Explanation

Area of triangle having vertices

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

$(x_3,y_3)$ is given by
Area

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

Given the vertices of triangle,

$A (a,b+c) ,B(a,b-c) ,C(-a,c)$

Therefore, area is given by

$=\dfrac { 1 }{ 2 } |\left[ a\left[ b-c-c \right] +a\left[ c-b-c \right] +(-a)\left[ b+c-b+c \right] \right]| \\ =\dfrac { 1 }{ 2 } |\left[ a(b-2c)+a(-b)-a(2c) \right]| \\ =\dfrac { 1 }{ 2 } |\left[ ab-2ac-ab-2ac \right]| =\left| \dfrac { -4ac }{ 2 } \right| =2ac$

If D (3, -1), E (2, 6) and F (5, 7) are the vettices of the sides of

$\Delta DEF$ , the area of triangle DEF is sq. units.

0%
$11$
0%
$22$
0%
$48$
0%
$50$

Explanation

Area of triangle =

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

Area of triangle

$DEF=|\dfrac{1}{2}(3(6-7)+2(7+1)+5(-1-6))|$

$=|\dfrac{1}{2}(-3+16-35)|$

$=11$ sq. Units.

If the straight line

$ax + by + p = 0$ and

$x\cos \alpha + y \sin \alpha = p$ enclosed an angle of

$\dfrac {\pi}{4}$ and the line

$x\sin \alpha - y \cos \alpha = 0$ meets them at the same point, then

$a^{2} + b^{2}$ is

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

$x\sin \alpha-y\cos\alpha =0$

$x=\dfrac{y\cos\alpha }{\sin\alpha}$

$x\cos\alpha+y\sin\alpha=p$

Put

$x=\dfrac{y\cos\alpha }{\sin\alpha}$ , we get

$\dfrac{y\cos^2\alpha}{\sin\alpha}+y\sin\alpha=p$

$y=p\sin\alpha$

Similiarly,

$x=p\cos\alpha$

$ax+by+p=0$

Put

$x=p\cos\alpha$ and

$y=p\sin\alpha$

$ap\cos\alpha+bp\sin\alpha+p=0$

$a\cos\alpha+b\sin\alpha=-1$ ---- Equation 1

Slope of

$ax+by+p=0$ ,

$m_1=-\dfrac{a}{b}$

Slope of

$x\cos\alpha+y\sin\alpha=p$ ,

$m_2=-\dfrac{\cos\alpha}{\sin\alpha}$

Also

$\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$

$\tan\dfrac{\pi}{4}=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$

$|m_1-m_2|=|1+m_1m_2|$

$\left|\dfrac{a}{b}-\dfrac{\cos\alpha }{\sin\alpha}\right|=\left|1+\dfrac{a\cos\alpha}{b\sin\alpha}\right|$

$|a\sin\alpha-b\cos\alpha|=|b\sin\alpha+a\cos\alpha|$

$|a\sin\alpha-b\cos\alpha|=1$ ---- Equation 2

Squaring and adding equation 1 and equation 2, we get

$(a^2\cos ^2\alpha +b^2\sin ^2\alpha +2ab\sin \alpha \cos \alpha)+(b^2\cos ^2\alpha +a^2\sin ^2\alpha -2ab\sin \alpha \cos \alpha)=1+1$

$a^2+b^2=2$

Option

$(C)$

The distance of the point

$A(a, b, c)$ from the x-axis is

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

Explanation

$\textbf{Step-1: Apply distance formula to get the required unknown.}$

$\text{The given point is}$

$(a,b,c)$

$\text{The distance of the point}$

$(a,b,c)$

$\text{from x-axis will be}$

$\text{the perpendicular distance from point}$

$(a,b,c)$

$\text{to x-axis whose co-ordinates will be}$

$(a,0,0)$

$\text{Now, we have to use the distance formula}$

$\text{Distance=}$

$\sqrt{(a-a)^{2}+(b-0)^{2}+(c-0)^{2}}=\sqrt{b^{2}+c^{2}}$

$\textbf{[Using distance formula]}$

$\textbf{Hence, the correct option is B}$

How many points

$(x, y)$ with integral co-ordinates are there whose distance from

$(1, 2)$ is two units?

0%
One
0%
Two
0%
Three
0%
Four

Explanation

$PQ = 2\Rightarrow PQ^{2} = 4$

$(x - 1)^{2} + (y - 2)^{2} = 4$
possibilities for integral coordinates

$\Rightarrow (X - 1)^{2} + (y -2)^{2} = 4 + 0$ or

$= 0 + 4$
Check

$(1) (x - 1)^{2} = 4\ (y - 2)^{2} = 0$

$\Rightarrow x - 1 = 2, -2\ ,y - 2 = 0$

$x = 3, -1\ ,y = 2$

$(3, 2)$ and

$(-1, 2)$
check

$(2) (x - 1)^{2} = 0\ ,(y - 2)^{2} = 4$

$x = 1\ ,y - 2 = 2, -2$

$x = 1\ ,y = 4, y = 0$

$(1, 4)$ and

$(1, 0)$
Total

$4$ integral coordinates are possible.

The distance between the points

$\left( a\cos { { 48 }^{ o } } ,0 \right)$ and

$\left( 0,a\cos { { 12 }^{ o } } \right)$ is

$d$ then

${ d }^{ 2 }{ a }^{ 2 }=$

0%
${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } +9 \right) }{ 4 }$
0%
${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } -1 \right) }{ 4 }$
0%
${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } -1 \right) }{ 8 }$
0%
${ a }^{ 2 }\cfrac { \left( \sqrt { 5 } +1 \right) }{ 8 }$

Explanation

Given

$A=(a\cos 48^o, 0)$ &

$B=(0, a \cos 12^o)$

$\Rightarrow d^2=(a\cos 48-0)^2+(0-a\cos 12)^2$

$=a^2[\cos^248+\cos^212]$

$\cos^2\theta =\dfrac{1+\cos 2\theta}{2}$

$\Rightarrow a^2\left[\left(\dfrac{1+\cos 96}{2}\right)+\left(\dfrac{1+\cos 24}{2}\right)\right]$

$\Rightarrow a^2\left[\dfrac{2+\cos 96+\cos 24}{2}\right]$

$\left[\cos A+\cos B=2\cos\left(\dfrac{A+B}{2}\right)\cos \left(\dfrac{A-B}{2}\right)\right]$

$\Rightarrow a^2\left[\dfrac{2+2\cos 60^o\cos 36^o}{2}\right]$

$\Rightarrow \dfrac{a^2}{2}\left[2+\not{2}\times \dfrac{1}{\not{2}}\cos 36^o\right]$

$d^2\Rightarrow \dfrac{a^2}{a}\left[2+\cos 36^o\right]$

We know

$\cos 36=\dfrac{\sqrt{5}+1}{4}$

$\therefore d^2a^2=\dfrac{a^4}{2}\left[2+\dfrac{\sqrt{5}+1}{4}\right]=\dfrac{a^4}{2}\left[\dfrac{\sqrt{5}+9}{4}\right]$

$\Rightarrow$ wrong options. [If d then]

$ABC$ is an isosceles triangle. If the coordinates of the base are

$B(1,3)$ and

$C(-2,7)$ . The vertex

$A$ can be

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$(1,6)$
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$(-1/2,5)$
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$(-7,1/6)$
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$(5/6,6)$

Explanation

ABC-Isosceles triangle.

Co-ordinates of base are

$B(1, 3)$ and

$C(-2, 7)$

vertex 'A' can be any point on perpendicular bisector of BC, but not 'D'.

Slope of BC

$=\dfrac{7-3}{-2-1}=-\dfrac{4}{3}$

Slope of Line(L)

$=\dfrac{-1}{\left(-\dfrac{4}{3}\right)}=\dfrac{3}{4}$

Mid-point of BC

$-$ D

$=\left(\dfrac{-1}{2}, 5\right)$

$L: 3x-4y=3\left(-\dfrac{1}{2}\right)-4(5)$

$3x-4y=-\dfrac{3}{2}-20$

$6x-8y=-3-40$

$6x-8y+43=0$

As, only

$(5/6, 6)$ satisfies 'L'.

So, 'A' can be

$\left(\dfrac{5}{6}, 6\right)$ .

The point

$P\left( x,y \right)$ is equidistant from the points

$Q\left( c+d,d-c \right)$ and

$R\left( c-d,c+d \right)$ then

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$cx=dy$
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$cx+dy=0$
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$dx=cy$
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$dx+cy=0$

Explanation

Point

$P(x,y)$ is the equidistant from

$Q(c+d,d-x)$ and

$R(c-d,c+d)$

$\therefore PQ=PR$

${ PQ }^{ 2 }={ PR }^{ 2 }$

${ \left[ x-\left( c+d \right) \right] }^{ 2 }+{ \left[ y-\left( d-c \right) \right] }^{ 2 }={ \left[ x-\left( c-d \right) \right] }^{ 2 }+{ \left[ y-\left( c+d \right) \right] }^{ 2 }$

${ x }^{ 2 }+{ \left( c+d \right) }^{ 2 }-2.x\left( c+d \right) +{ y }^{ 2 }+{ \left( d-c \right) }^{ 2 }-2.y(d-c)={ x }^{ 2 }+{ \left( c-d \right) }^{ 2 }-2.x.\left( c-d \right) +{ y }^{ 2 }+{ \left( c+d \right) }^{ 2 }+2.y(c+d)$
Or

$x(c+d)+y(d-c)=x(c-d)+y(c+d)$
Or

$(c+d)(x-y)=(c-d)(x+y)$
Or

$cx-cy+dc-dy=cx+cy-dx-dy$
Or

$2dx=2cy$
Or

$dx=cy$

$\therefore$ Answer

$dx=cy$

The area of triangle formed by the lines

$x+y-3=0$ ,

$x-3y+9=0$ and

$3x-2y+1=0$ is:

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$\cfrac { 16 }{ 7 }$ sq. units
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$\cfrac { 10 }{ 7 }$ sq. units
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$4$ sq. units
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$9$ sq. units

Explanation

$L_1:x+y-3=0$

$L_2:x-3y+9=0$

$L_3:3x-2y+1=0$

$A$ is intersection point of

$L_1$ and

$L_2$ which is

$(0,3)$ on solving

$L_1$ and

$L_2$ .

$B$ is intersection point of

$L_2$ and

$L_3$ which is

$(1,2)$ on solving

$L_2$ and

$L_3$ .

$C$ is intersection point of

$L_3$ and

$L_1$ which is

$(\dfrac{15}{7},\dfrac{26}{7})$ on solving

$L_3$ and

$L_1$ .

Now, area of

$\Delta ABC=\dfrac{1}{2}|(x_1y_2-x_2y_1)+(x_2y_3-x_3y_2)+(x_3y_1-x_1y_3)|$

$=\dfrac{1}{2}|(0-3)+(\dfrac{26}{7}-\dfrac{30}{7})+(\dfrac{45}{7}-0)|$

$=\dfrac{1}{2}|-3-\dfrac{4}{7}+\dfrac{45}{7}|=\dfrac{1}{2}\times \dfrac{20}{7}$

$=\dfrac{10}{7} sq. units$

A circle passes through the points

$(2, 3)$ and

$(4, 5)$ . If its centre lies on the line,

$y - 4x + 3 = 0$ , then its radius is equal to

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$\sqrt {5}$
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$1$
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$\sqrt {2}$
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$2$

Explanation

Equation of the line through the given points is

$y-3 = x-2 \Rightarrow x-y+1=0$ .

Equation of the perpendicular line through the midpoint

$(3,4)$ is

$x+y-7=0$ .

This intersects the given line at the center of the circle. So, the center of the circle is found to be

$(2,5)$ .

Clearly, the radius is then

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} =$

$\sqrt{(2-2)^2+(3-5)^2} = 2$ units.

So the answer is option D.

The angle between the pair of lines whose equation is

$4{ x }^{ 2 }+10xy+m{ y }^{ 2 }+5x+10y=0$ is

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$\tan ^{ -1 }{ \left( \dfrac { 3 }{ 8 } \right) }$
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$\tan ^{ -1 }{ \dfrac {2 \sqrt { 25-4m } }{ m+4 } }$
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$\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) }$
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$\tan ^{ -1 }{ \dfrac { \sqrt { 25-4m } }{ m+4 } }$

Explanation

The angle b/w 2 line represented by

$ax^2+2hxy+by^2+2gx+2fy+c=0$

$\tan\theta=\Bigg|\dfrac{2\sqrt{h^2-ab}}{a+b}\Bigg|$

Given the equation of line i.e.

$4x^2+10xy+my^2+5x+10y=0$

Compare above equation with standard equation

$a=4,h=5,b=m,g=\dfrac{5}{2},f=5,c=0$

Hence

$\theta=\tan^{-1}\dfrac{2\sqrt{25-4m}}{m+4}$

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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