MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 9

The points

$(3,0), (6,4)$ and

$(-1,3)$ are vertices of a right-angled triangle. Show that it is an isosceles triangle.

0%
True
0%
False

Explanation

Given,

$A(3,0),B(6,4),C(-1,3)$

$AC=\sqrt{(-1-3)^2+(3-0)^2}=5$

$AB=\sqrt{(6-3)^2+(4-0)^2}=5$

$BC=\sqrt{(-1-6)^2+(3-4)^2}=5\sqrt{2}$

Since 2 sides are equal, these points are the vertices of isosceles triangle.

If the points

$P(a,-11), Q(5,b), R(2,15)$ and

$S(1,1)$ are the vertices of a parallelogram

$ABCD$ , find the values of

$a$ and

$b$ .

0%
$a=5,b=3$
0%
$a=4,b=10$
0%
$a=4,b=3$
0%
None of these

Explanation

Diagonals of a parallelogram bisect each other.

So the mid point of diagonal

$PR$ and mid point of diagonal

$QS$ coincide.

We have mid point formula,

$P(X,Y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

It implies that

$\dfrac{a+2}{2}=\dfrac{5+1}{2}$

$a+2=6$

$a=4$

Similarly,

$\dfrac{b+1}{2}=\dfrac{-11+15}{2}$

$b+1=4$

$b=3$

1350138_973778_ans_dc4f22d60277434e9f5de9bf034b1fe5.png

State true or false

Points

$( -2 , -1 ) , (1 , 0 ) , ( 4 , 3)$ and

$( 1 , 2)$ are the vertex of the parallelogram

0%
True
0%
False

Explanation

Let the given points be

$A,B,C$ and

$D$ respectively.Then,

Coordinates of the mid-point of

$AC$ are

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

Coordinates of the mid-point of

$BD$ are

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

Thus,

$AC$ and

$BD$ have the same mid-point.

Hence,

$ABCD$ is a parallelogram.

The area of the triangle with vertices at

$(-4, 1), (1, 2)(4, -3)$ is

0%
$14$
0%
$16$
0%
$15$
0%
None of these

Prove that The lines

$ax + by + c=0$ and

$Ax + By + C = 0$ are perpendicular if

$aA + bB = 0$ .

0%
True
0%
False

The coordinates of the point where the line joining

$P(3, 4, 1)$ and

$Q(5, 1, 6)$ crosses the xy-plane are:

0%
$\left( { - {{13} \over 5}, - {{23} \over 5},0} \right)$
0%
$\left( {{{13} \over 5},{{23} \over 5},0} \right)$
0%
$\left( {{{13} \over 5}, - {{23} \over 5},0} \right)$
0%
$\left( { - {{13} \over 5},{{23} \over 5},0} \right)$

Explanation

$P(3,4,1)$

$Q(5,1,6)$

Equation of PQ

$\cfrac { x-3 }{ 5-3 } =\cfrac { y-4 }{ 1-4 } =\cfrac { z-1 }{ 6-1 } \\ \cfrac { x-3 }{ 2 } =\cfrac { y-4 }{ -3 } =\cfrac { z-1 }{ 5 } \\ In\quad xy\quad plane\quad z=0\\ \therefore \cfrac { z-1 }{ 5 } =\cfrac { -1 }{ 5 } \\ \cfrac { x-3 }{ 2 } =\cfrac { -1 }{ 5 } \\ x=3-\cfrac { 2 }{ 5 } =\cfrac { 13 }{ 5 } \\ \cfrac { y-4 }{ -3 } =\cfrac { -1 }{ 5 } \\ y=4+\cfrac { 3 }{ 5 } =\cfrac { 23 }{ 5 } \\ \therefore Coordinates\quad are\left( \cfrac { 13 }{ 5 } ,\cfrac { 23 }{ 5 } ,0 \right) \\ \\$

The area of the triangle whose vertices are (3,8), (-4,2) and (5,-1) is

0%
$75 \ sq. units$
0%
$37.5 \ sq. units$
0%
$45 \ sq. units$
0%
$22.5 \ sq. units$

Explanation

Let

$A(3,8), B(-4,2),C(5,-1)$ be the vertices of the given

$\triangle ABC$ .

Then,

$(x_{1}=3,y_{1}=8),(x_{2}=-4,y_{2}=2),(x_{3}=5,y_{3}=-1)$

Area of

$\triangle ABC$ =

$\dfrac{1}{2}|[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]|$

$=\dfrac{1}{2}|3[2-(-1)]-4(-1-8)+5(8-2)|$

$=\dfrac{1}{2}|9+36+30|=\dfrac{75}{2}=37.5 \ sq. units$

$(3,2)$ and

$(-3,2)$ are two vertices of an equilateral triangle which contains within it the origin, what are the coordinates of the third vertex ?

0%
$(2,2\sqrt{2})$
0%
$(1,2\sqrt{2})$
0%
$(0,-3\sqrt{3})$
0%
$None\ of\ these$

Explanation

Let $\Delta ABC$ be the equilateral triangle, where $AB=BC=AC$ .

Vertices of this triangle be,

$A\left( 3,2 \right),B\left( -3,2 \right)$ and $C\left( x,y \right)$ .

The mid-point of the side $AB$ is $M\left( 0,2 \right)$ .

$AB=\sqrt{{{\left( 3+3 \right)}^{2}}+{{\left( 2-2 \right)}^{2}}}$

$AB=6\ units$

Therefore,

$\Rightarrow AB=BC=AC=6\ units$

Also,

$\Rightarrow AM=3\ units$

As two vertices are $A\left( 3,2 \right)$ and $B\left( -3,2 \right)$ , so third vertex will be at $y-$ axis.

Therefore,

$\Rightarrow C\left( 0,y \right)$

It will be located below the origin as the triangle contains the origin.

Now,

${{y}^{2}}={{\left( AC \right)}^{2}}-{{\left( AM \right)}^{2}}$

${{y}^{2}}=36-9=27$

$y=3\sqrt{3}\ units$

Therefore,

$\Rightarrow C\left( 0,-3\sqrt{3} \right)$

Hence, this is the required point.

The curve

$y=ax^3+bx^2+cx+5$ touches the x-axis at

$P(-2,0)$ and cuts the y-axis at a point

$Q$ where its gradient is

$3$ . Then the value of

${a,b,c}$ is

0%
$A=\dfrac{9}{2}$ , $B=\dfrac{1}{2}$ , $C=\dfrac{3}{5}$
0%
$A=\dfrac{-1}{2}$ , $B=\dfrac{-3}{4}$ , $C=3$
0%
$A=\dfrac{9}{2}$ , $B=\dfrac{-3}{4}$ , $C=3$
0%
$A=\dfrac{-9}{2}$ , $B=\dfrac{1}{2}$ , $C=\dfrac{-3}{5}$

Explanation

$y=ax^{3}+bx^{2}+cx+5$

$P(-2, 0)$ at

$y$ axis

differentiate

$y$ wrt

$x$

$y=5 @\phi (0, 5) m=3$

$\left(\dfrac{dy}{dx}\right)_{@\phi}=3ax^{2}+2bx+c$

$\left(\dfrac{dy}{dx}\right)_{@(0, 5)}=3a(0)^{2}+2b(0)+c=3$

$c=3$

$\left(\dfrac{dy}{dx}\right)_{@ P=(-2, 0)}=3a(-2)^{2}+2b(-2)+3=0$

$12a-4b+3=0 .... (1)$

$-8a+4b-1=0 ..... (2)$

$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$

$a=-1/2 b=-3/4$

Firdinates of A, B and C are $(2,1)$ , $(x,0)$ and $(-2,-1)$ respectively.

0%
$1$
0%
$2$
0%
$-2$
0%
$0$

Explanation

Distance between two points

$=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Given

$AB=BC$

$\sqrt{(2-x)^{2}+(1)^{2}}=\sqrt{(x+2)^{2}+(1)^{2}}$

Squaring both sides

$(2-x)^{2}+(1)=(x+2)^{2}+(1)$

$4+x^{2}-4x=x^{2}+4+4x$

$x=0$

Find the value of

$x_i$ , if the distance between the points

$(x_i, 2)$ and

$(3, 4)$ is

$8$ .

0%
$3\pm 2\sqrt{15}$
0%
$3\pm \sqrt{15}$
0%
$2\pm 3\sqrt{15}$
0%
$2\pm \sqrt{15}$

Explanation

we re given

$( x_1 , y_1 )$ = ( xi, 2) &

$B ( x_2,y_2)=(3,4)$

let d is the distance between point A & B . d= 8

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

$8 = \sqrt { (3 -x_i )^2 +(4-2)^2 }$

taking square on both side,

$64 = (3 -x_i)^2 + (4-2)^2$

$64 = 9 - 6x_i +(x_i)^2 +(2)^2$

$(x_i)^2-6x_i - 51 = 0$

now, discriminant

$D = \sqrt { b^2 - 4ac }$

here

$b = -6$ ,

$a = 1$ ,

$c= -51$

so,

$d= \sqrt { (-6)^2 -4(1)(-51) } = \sqrt {36+ 204 }$

$d= \sqrt {240} > 0$

so,

$\displaystyle x_i = \frac { -b \pm \sqrt d }{2a} = \frac { -(-6) \pm \sqrt {240} }{ 2(1) } = \frac { 6 \pm 4 \sqrt {15} }{2}$

$x_i = 3 \pm 2 \sqrt {15}$

The coordinates of a point on Y-axis which are at a distance of

$5\sqrt 2$ from the point

$P(3,-2,5)$ is -

0%
$(0,6,0)$ or $(0,-2,0)$
0%
$(0,-6,0)$ or $(0,2,0)$
0%
$(0,4,0)$ or $(0,-4,0)$
0%
$(0,5,0)$ or $(0,-5,0)$

Explanation

${\rm{since point is an y - axis}}$

${\rm{its x and z coordinate is 0}}$

${\rm{Let pt}}{\rm{. on y - axis be A (0,a,0)}}$

${\rm{PA = }}\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}}$

$5\sqrt2 = \sqrt {{{( - 3)}^2} + {{(a + 2)}^2} + {{(0 - 5)}^2}}$

$5\sqrt2 = \sqrt {a + {a^2} + 4 + 4a + 25}$

$5\sqrt2 = \sqrt {{a^2} + 4a + 38}$

$25 \times 2 = {a^2} + 4a + 38$

${a^2} + 4a - 12 = 0$

${a^2} + 6a - 2(a + b) = 0$

$a = 2,a = - 6$

$co - ordinate{\rm{ of point (0,2,0)and(0, - 6,0)}}$

C is a point on the line segment joining A(-3, 4) and B (2, 1) such that AC = 2 BC then coordinates of C are

0%
$\left( {\dfrac{1}{3},2} \right)$
0%
$\left( {2,\dfrac{1}{3}} \right)$
0%
$\left( {2,7} \right)$
0%
$\left( {7,2} \right)$

Explanation

AC = 2 BC

${{AC} \over {BC}} = {2 \over 1}$

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

$y = {{2 \times 1 + 1 \times 4} \over {2 + 1}} = {{2 + 4} \over 3} = 2$

$C\left( {{1 \over 3},2} \right)$

Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are

$(0.-1), (2, 1) and (0, 3)$

0%
$4$
0%
$8$
0%
$1$
0%
$2$

Explanation

Area of triangle having vertices

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

$(x_3,y_3)$ is given by
Area

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

Also, we know

$\\Area\>of\>triangle\>=4\times\>of\>triangle\>formed\>using\>mid-point\>\\=4\times(\frac{1}{2})[x_1(y_2-y_3)+x_2(y_3+y_1)+x_3(y_1-y_2)]\\=2[0+2(3-1)+0]=8sq\>unit$

$Q(0, 1)$ is equidistant from

$P(5, -3)$ and

$R(x, 6)$ , find the values of x. Also find the distances QR and PQ.

0%
PQ=QR= $\sqrt { 41}$
0%
PQ=20, QR= $\sqrt { 29 }$
0%
PQ= $\sqrt { 29 }$ , QR= 20
0%
PQ $=\sqrt { 19 }$ , QR = 21

Explanation

Q is equidistant from P and R

=>PQ=RQ

$P{ Q }^{ 2 }=R{ Q }^{ 2 }$

$(5)^{ 2 }+(-4)^{ 2 }={ x }^{ 2 }+{ (5) }^{ 2 }\\ =>{ x }^{ 2 }=16\\ =>x=\pm 4$

Now PQ=QR=

$\sqrt { 16+25 } =\sqrt { 41 }$

What is the y intercept of the line that is parallel to

$y=3x,$ and which bisects the area of rectangle with corners at

$(0,0), (4,0) ,(4,2)$ and

$(0,2)$ ?

0%
$-7$
0%
$-6$
0%
$-5$
0%
$-4$

Explanation

Rectangle's midpoints are

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

Slope line y

$=$ 3x will be :

${m_1} = 3$

Parallel line will be

${m_2} = 3$

Equation of line passing through (2 , 1)

$y - {y_1} = m(x - {x_1})$

$(y - 1) = 3(x - 2)$

$y - 1 = 3x - 6$

$y = 3x - 5$

Hence on comparing

Y- intercept

$=$ - 5

Find the area of the triangle whose vertices are

$(-5, -1), (3, -5), (5, 2)$

0%
$31 sq$ $units$
0%
$32sq$ $units$
0%
$33 sq$ $units$
0%
$\text{no triangle can be formed}$

Explanation

Area of a triangle whose vertices are $A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)$ is given as,

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

Let $A$ be the required area

$A=(\dfrac{1}{2})[(-5)(-5-2)+3(2+1)+5(-1+5)]\\ (\dfrac{1}{2})[35+9+20]\\=32sq\>unit$

So, the correct option is (B)

An aeroplane flying horizontally

$1$ , above the ground is observed at an elevation of

$60$ and after

$10$ seconds the elevation is observed to be

$30$ . The uniform speed of the aeroplane in

$/h$ is-

0%
$240$
0%
$240\sqrt { 3 }$
0%
$60\sqrt { 3 }$
0%
None of these

Explanation

$t=0$ plane is at A and at

$t=10$ , the plane is at B. Distance covered

$=AB$

$\triangle AHO$

$\tan { 60° } =\cfrac { OH }{ AH }$

$AH=\cfrac { (1000) }{ \sqrt { 3 } } m$

$\triangle BHO$

$\tan { 30° } =\cfrac { OH }{ BH }$

$BH=\sqrt { 3 } \left( 1000 \right) m$

$\therefore AB=BH-AH=\left( \sqrt { 3 } -\cfrac { 1 }{ \sqrt { 3 } } \right) \left( 1000 \right) m$

$=\cfrac { 2 }{ \sqrt { 3 } } .1000m$

Speed

$=\cfrac { Distance }{ Time }$

Speed

$=\cfrac { 2 }{ \sqrt { 3 } } \times \cfrac { 1000 }{ 10 } ㎧$

$=\cfrac { 200 }{ \sqrt { 3 } } ㎧$

$=\cfrac { 200 }{ \sqrt { 3 } } \times \cfrac { 18 }{ 5 } ㎞{ h }^{ -1 }$

$=40\times 6\sqrt { 3 }$

Speed

$=240\sqrt { 3 } ㎞{ h }^{ -1 }$

982935_1075136_ans_39f995ba53554339a1d6c1f5f0bccff8.png

The points

$A(a, 0), B(0, b), C(c, 0)$ &

$D(0, d)$ are such that

$ac=bd$ & a, b, c, d are all non-zero. Then the points.

0%
Form a parallelogram
0%
Do not lie on a circle
0%
Form a trapezium
0%
Are concyclic

Explanation

The points

$A(a,0),B(0,b),C(c,0)$ and

$D(0,d)$ are such that

$ac=bd$ and

$a, b, c, d$ are all non-zero.

$ac=bd$ [ Given ]

$\Rightarrow$

$AO\times OC=OB\times OD$

This is true in case of circle and two secants.

$\therefore$

$A,B,C$ and

$D$ lie on a circle.

$\therefore$ The given points are concyclic.

1308701_1070365_ans_9acff6a827844d66bec5c2a1a1f8cdfd.png

Find the coordinates of the point equidistant from the points

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

$C(7, -6)$ .

0%
$(3, -3)$
0%
$(-3, 3)$
0%
$(-2, -3)$
0%
$(-3, -3)$

Explanation

Let us consider the co-ordinate

$(x,y)$ which is the point equidistant from the points

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

$C(7,-6).$

So,

$d_{1}= \sqrt{(x-(-2))^{2}+ (y-(-3))^{2}} = \sqrt{(x+2)^{2}+ (y+3)^{2}}$

$d_{2}= \sqrt{(x-(-1))^{2}+ (y-0)^{2}} = \sqrt{(X+1)^{2} +y^{2}}$ &

$d_{3}= \sqrt{(x-7)^{2}+ (y-(-6))^{2}}^{2} = \sqrt{(x-7)^{2}+ (y+6)^{2}}$

here

$d_{1}= d_{2}= d_{3}$

Let compare

$d_{1}$ &

$d_{2}$ . we get,

$\sqrt{(x+2)^{2} + (y+3)^{2}}= \sqrt{(x+1)^{2}}+y^{2}$

$\therefore (x+2)^{2}+ (y+3)^{2}= (x+1)^{2}+ y^{2}$

$\therefore x^{2}+4x+4+ y^{2}+ 6y+ 9 = x^{2}+ 2x+1+y^{2}$

$\therefore 2x+ 12+ 6y= 0$

$\therefore x+ 3y+ 6 = 0 \rightarrow (1)$

Now, compare

$d_{2}$ &

$d_{3}$ . we get.

$\sqrt{(x+1)^{2}+y^{2}}= \sqrt{(x-7)^{2}+ (y+6)^{2}}$

$x^{2}+2x+1+y^{2}= x^{2}-14x+49+y^{2}+12y+36$

$16x-12y= 84$

$4 (4x-3y)= 84$

$4x-3y- 21= 0 \rightarrow (2)$

From

$(1)$ &

$(2)\ 5x- 15 = 0 \Rightarrow x=3$

From

$(1) \Rightarrow y= \dfrac{-x-6}{3}= \dfrac{-3-6}{3}=\dfrac{-9}{3}= -3$

So,

$(x,y)= (3,-3)$

Distance between A(x, y) and B(-4, 7) is

$\sqrt{41}.$ Find x and y, if A's ordinate is thrice of its abscissa.

0%
$(\frac{-12}{5},\frac{36}{5})$
0%
$(1,\dfrac{12}{5})$
0%
$\text{both a and b}$
0%
none

Explanation

$\\(x+4)^2(y-7)^2=41\\and y=3x \\\therefore (x+4)^2+(3x-7)^2=41\\x^2+8x+16+9x^2+49-42x=41\\10x^2-34x+24=0\\5x^2-17x+12\\5x^2-5x-12x+12=0\\5x(x-1)-12(x-1)=0\\or (x-1)(5x-12)=0\\\therefore x=1 and x=(\frac{12}{5})$

The ends of the base of an isosceles triangle are at

$(2a, 0)$ and

$(0, a)$ and one side is parallel to Y-axis. The equation of the other side is?

0%
$x+2y-a=0$
0%
$x+2y=2a$
0%
$3x+4y-4a=0$
0%
$3x-4y+4a=0$

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

$C(-2,-4)$ are three points, then the angle between

$BA$ and

$BC$ is:

0%
$\tan ^{ -1 }{ \left( \cfrac { 3 }{ 2 } \right) }$
0%
$\tan ^{ -1 }{ \left( \cfrac { 2 }{ 3 } \right) }$
0%
$\tan ^{ -1 }{ \left( \cfrac { 7 }{ 4 } \right) }$
0%
$none\ of\ these$

Explanation

Slope of line joining

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

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

First Consider the side

$BA$ . Let

$m_1$ be the slope of

$BA$

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

$\Rightarrow m_1=\dfrac{2}{4}$

First Consider the side

$BC$ . Let

$m_2$ be the slope of

$BC$

$\Rightarrow m_2=\dfrac{3+4}{2+2}$

$\Rightarrow m_2=\dfrac{7}{4}$

Angle between

$BA, BC$ is given by

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

Substituting the values of

$m_1, m_2$ we get

$\Rightarrow \tan\theta=\dfrac{\dfrac{5}{4}}{1+\dfrac{14}{16}}$

On solving we get

$\Rightarrow \tan\theta=\dfrac{2}{3}$

$\Rightarrow \theta=\tan^{-1}(\dfrac{2}{3})$

A straight line L through the point

$(3,-2)$ is inclined at an angle

$60$ to the line

$\sqrt { 3 } x+y=1$ . If L also intersects the x-axis, the equation of L is-

0%
$y+\sqrt { 3 }x+2-3\sqrt { 3 }=0$
0%
$y-\sqrt { 3 }x+2+3\sqrt { 3 }=0$
0%
$\sqrt { 3 } y-x+3+2\sqrt { 3 } =0$
0%
$\sqrt { 3 } y+x-3+2\sqrt { 3 } =0$

Explanation

$\sqrt{3}x+y=1\implies m=\tan \theta =-\sqrt{3}\implies$ this line makes an angle of

$120^{\circ}$ with x-axis.

Since

$L$ makes an angle

$60^{\circ}$ with this line.

$\implies L$ must be either x- axis

$($ which is not correct because given

$L$ intersect the x-axis

$)$ or a line making an angle

$60^{\circ}$ with the x-axis.

$\implies m=\tan 60^{\circ}=\sqrt{3}$

$L:y=m{x}+c$

$L:y=\sqrt{3}x+c$

$L$ passes through

$(3,-2)\implies -2=3\sqrt{3}+c\implies c=-2-3\sqrt{3}$

$L:y-\sqrt{3}x+2+3\sqrt{3}=0$

A straight line is drawn through the point

$p\ (2,3)$ and is inclined at an angle of

$30^{o}$ with the

$x-$ axis, the co-ordinates of two points on it at a distance of

$4$ from

$p$ is/are

0%
$\left(2+2\sqrt{3},5\right)$ , $\left(2-2\sqrt{3},1\right)$
0%
$\left(2+2\sqrt{3},5\right)$ , $\left(2+2\sqrt{3},1\right)$
0%
$\left(2-2\sqrt{3},5\right)$ , $\left(2-2\sqrt{3},1\right)$
0%
$none\ of\ these$

Explanation

The parametric points of a line is given as

$(x_1\pm r\cos \theta,y_1\pm r\sin \theta)\\(2\pm 4\cos 30,3\pm 4 \sin 30)\\\left(2\pm 4\dfrac{\sqrt3}{2},3\pm 4\dfrac 12\right)\\(2\pm2\sqrt3,3\pm 2)\\(2+2\sqrt3,5),(2-2\sqrt3,1)$

The points given are

$(1, 1)$ ,

$(-2, 7)$ and

$(3, 3)$ .Find distance between the points.

0%
$\sqrt45,$ $7,$ $\sqrt{8}$
0%
$\sqrt{10},7,8$
0%
$\sqrt{3},1,2$
0%
None of the above

Explanation

Let the points be

$A(1,1)$ ,

$B(-2,7)$ ,

$C(3,3)$

Distance of

$AB=\sqrt { { \left( 2-1 \right) }^{ 2 }+{ \left( 7-1 \right) }^{ 2 } }$

$=\sqrt { 9+36 } =\sqrt { 45 }$

Distance of

$BC=\sqrt { { \left( 3+2 \right) }^{ 2 }+{ \left( 3-7 \right) }^{ 2 } }$

$=\sqrt { 25+16 } =\sqrt { 49 } =7$

Distance of

$AC=\sqrt { { \left( 3-1 \right) }^{ 2 }+{ \left( 3-1 \right) }^{ 2 } } =\sqrt { 8 }$

The points

$(a, b), (-a, -b), (b\sqrt{3}, -a\sqrt{3})$ are the vertices of a triangle which is

0%
Isosceles
0%
Equilateral
0%
Right angled
0%
Scalene

A line passing through

$(3,4)$ meets the axis

$\overline {OX}$ and

$\overline {OY}$ at

$A$ and

$B$ respectively. Find the minimum area of triangle OAB.

0%
$12$
0%
$10$
0%
$24$
0%
$6$

If the vertices of a triangle are

$(1,2),(4,-6)$ and

$(3,5)$ , then its area is

0%
$\cfrac{23}{2}$ sq. units
0%
$\cfrac{25}{2}$ sq. units
0%
$12$ sq.unit
0%
none of these

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, area of required triangle is given by

Area

$= \dfrac{1}{2} \times [ 1 (-6 - 5) + 4(5 - 2) + 3(2 + 6) ]$

$=\dfrac{1}{2}[-11+12+24]$

$=\dfrac{1}{2}\times25$

$=\dfrac{25}{2}\ sq.unit$

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

$B(1,3)$ and

$C(-2,7)$ , the coordinates of vertex

$A$ can be

0%
$(1,6)$
0%
$(-1/2,5)$
0%
$(5/6,6)$
0%
none of these

Explanation

$AB=AC$

$\sqrt{(x-1)^2+(y-3)^2}=\sqrt{(x-(-2))^2+(y-7)^2}$

$(x-1)^2+(y-3)^2=(x+2)^2+(y-7)^2$

$=2x+1-6y+9=4x+4-14y+49$

$\Rightarrow 6x-8y+43=0$

By substituting

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

$\Rightarrow \not{6}\times \dfrac{5}{\not{6}}-8\times 6+43$

$\Rightarrow 0$

Hence

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

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 9 - 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 9 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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