MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 3

Which of the following are the co-ordinates of the centre of the circle that passes through

$P(6, 6), Q(3, 7)$ and

$R(3, 3)$ ?

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

Explanation

Let centre be

$A(x,y)$

$AP^2 = AQ^2$ (Radii of same circle)

$(x-6)^2 + (y + 6)^2 = (x-3)^2 + (y + 7)^2$

$\Rightarrow 3x+y = 7$ ...(1)

Similarly

$AQ^2 = AR^2$

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

$\Rightarrow y = -2$ ...(2)

From (1) and (2)

$x = 3$

Find the area of triangle having vertices are

$A (3, 1), B (12, 2)$ and

$C (0, 2)$ .

0%
$4$
0%
$6$
0%
$12$
0%
$18$

Explanation

The given points are $A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 3,1 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 12,2 \right) \& C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 0,2 \right)$

Let us obtain $ar\Delta ABC$ by applying the area formula.

$ar\Delta =\dfrac { 1 }{ 2 } \left\{ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right\}$

$\Rightarrow ar\Delta =\dfrac { 1 }{ 2 } \left\{ 3\left( 2-2 \right) +12\left( 2-1 \right) +0\left( 1-2 \right) \right\}$ units $=6$ units

$\therefore ar\Delta =6$ units

Hence, option B.

What type of a quadrilateral do the points

$A (2, 2), B (7, 3), C (11, 1)$ and

$D (6, 6)$ taken in that order, form?

0%
Scalene quadrilateral
0%
Square
0%
Rectangle
0%
Rhombus

Explanation

Given that the point

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,2 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 7,3 \right) , C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 11, 1 \right)$ and

$D\left( { x }_{ 4 },{ y }_{ 4 } \right) =\left( 6,6 \right)$ form a quadrilateral

$ABCD$ .

Its type will be (i) a scalene one, if

$AC\neq BD$ , (ii) a rectangle, if

$AC=BD$ .

We apply the distance formula, to get

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

$\therefore AC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-{ y }_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( 11-2 \right) }^{ 2 }+{ (1-2) }^{ 2 } }$ units

$=\sqrt { 82 }$ units and

$BD=\sqrt { { \left( { x }_{ 4 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 4 }-{ y }_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 6-7 \right) }^{ 2 }+{ (3-6) }^{ 2 } }$ units

$=\sqrt { 10 }$ units

So,

$AC\neq BD$

$\therefore ABCD$ is a scalene quadrilateral.

Also,

$AB=\sqrt { { \left( 7-2\right) }^{ 2 }+{ (3-2) }^{ 2 } } =\sqrt{26}$ units

$BC=\sqrt { { \left( 11-7\right) }^{ 2 }+{ (1-3) }^{ 2 } } =\sqrt{20}$ units

Thus, the sides are not equal. So, they does not represent a rhombus or a square.

Hence, only option A is correct.

The area of a triangle with vertices

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

$(c, a + b)$ is

0%
$a$
0%
$b$
0%
$c$
0%
$0$

Explanation

The area of

$\triangle ABC$ with vertices

$A\equiv(x_1, y_1)$ ,

$B\equiv(x_2, y_2)$ and

$C\equiv(x_3, y_3)$ is given as,

$A(\triangle ABC) = \left|\dfrac12[x_1(y_3-y_2) +x_2(y_1 - y_3) + x_3(y_2-y_1) ]\right|$

In this problem,

$A(\triangle ABC) = \left|\dfrac12[a(a+b-(c+a)) +b(b+c - (a+b)) + c(c+a-(b+c)) ]\right|$

$\therefore A(\triangle ABC) = \left|\dfrac12[ab-ac + bc-ba + ca-cb]\right|$

$\therefore A(\triangle ABC) = 0$

Hence, the correct Option is D.

The line

$3x-4y+8=0$ is rotated through an angle

$\displaystyle \frac { \pi }{ 4 }$ in the clockwise direction about the point

$(0,2)$ . The equation of the line in its new position is

0%
$7y+x-14=0$
0%
$7y-x-14=0$
0%
$7y+x-2=0$
0%
$7y-x=0$

Explanation

$\displaystyle \tan { \theta } =\frac { 3 }{ 4 } ,\phi =\theta +\frac { 3\pi }{ 4 }$

$\displaystyle m=\tan { \phi } =\tan { \left( \theta +\frac { 3\pi }{ 4 } \right) } =\frac { \tan { \theta } -1 }{ 1+\tan { \theta } } =\frac { \frac { 3 }{ 4 } -1 }{ 1+\frac { 3 }{ 4 } } =-\frac { 1 }{ 7 }$
Equation of

$AC$ is

$\displaystyle y-2=-\frac { 1 }{ 7 } \left( x-0 \right) \Rightarrow x+7y-14=0$

Find the area of triangle having vertices

$A (6, 10), B (4, 5)$ and

$C (3, 8)$ .

0%
$5$
0%
$5.5$
0%
$10$
0%
$11$

Explanation

The point are

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 6,10 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 4,5 \right)$ and

$C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 3,8 \right)$ .

$ar\Delta ABC=\dfrac { 1 }{ 2 } \left\{ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right\}$

$\Rightarrow ar\Delta ABC=\dfrac { 1 }{ 2 } \left\{ 6\left( 5-8 \right) +4\left( 8-10 \right) +3\left( 10-5 \right) \right\}$ sq. units

$\Rightarrow ar\Delta ABC=5.5$ sq. units

Plot

$(3, 0), (5, 0)$ and

$(0, 4)$ on cartesian plane. Name the figure formed by joining these points and find its area.

0%
Figure formed is a Rectangle and Area $= 52$ sq. unit
0%
Figure formed is a triangle and Area $= 4$ sq. unit
0%
Figure formed is a square and Area $= 40$ sq. unit
0%
None of these

Explanation

The figure is a triangle.
We know that the area of the triangle whose vertices are

$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$ and

$(x_{3},y_{3})$ is

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

Therefore,

Area

$= \dfrac{1}{2} [3(0-4)+5(4-0)+0] = 4$ square units.

The area of a triangle with vertices

$A (3, 0), B (7, 0)$ and

$C (8, 4)$ is

0%
$14$
0%
$28$
0%
$8$
0%
$6$

Explanation

The area of the

$\Delta ABC$ with vertices

$A\left( { x }_{ 1 },{ y }_{ 1 } \right)=(3,0), B\left( { x }_{ 2 },{ y }_{ 2 } \right)=(7,0)\ \&\ C\left( { x }_{ 3 },{ y }_{ 3 } \right)=(8,4)$ is

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

$=\dfrac { 1 }{ 2 } \left\{ 3\left( 0-4 \right) +7\left( 4-0 \right) +8\left( 0-0 \right) \right\}$ sq.units

$=8$ sq. units.

Hence, option C.

If the perpendicular distance of a point

$P$ from the

$x$ -axis is

$5$ units and the foot of the perpendicular lines on the negative direction of

$x$ -axis, then the point P has?

0%
$x$ coordinate $= 5$
0%
$y$ coordinate $= 5$ only
0%
$y$ coordinate $= - 5$ only
0%
$y$ coordinate $= 5$ or $-5$

Explanation

$The\quad foot\quad of\quad the\quad perpendicular\quad from\quad the\quad point\quad P\quad to\\ the\quad X-axis\quad is\quad to\quad the\quad negative\quad direction\quad of\quad X-axis.\\ \therefore \quad Its\quad coordinate\quad is\quad (-x,0)\\ The\quad length\quad of\quad the\quad perpendicular\quad is\quad 5\quad units.\\ So\quad the\quad ordinate\quad of\quad P\quad will\quad be\quad 0\pm 5=+5\quad or\quad -5.\\ Ans-\quad Option\quad D$

The points

$A (2, 9), B (a, 5)$ and

$C (5, 5)$ are the vertices of a triangle

$ABC$ right angled at

$B$ . Find the values of

$a$ and hence the area of

$\Delta$

$ABC$ .

0%
$a = -1$ , Area $=$ $15$ sq. unit
0%
$a = 2$ , Area $=$ $6$ sq. unit
0%
$a = 0$ , Area $=$ $8$ sq. unit
0%
$a = 1$ , Area $=$ $12$ sq. unit

Explanation

Given that the vertices of the

$\Delta ABC$ are

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,9 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( a,5 \right)$ and

$C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 5,5 \right) and \angle B={ 90 }^{ o }$ .

So,

$AC$ is the hypotenuse.

$\therefore$ by Pythagoras theorem, we have

${ AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$ .........(i)

Now, by distance formula

$d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } }$

So,

$AB=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 2-a \right) }^{ 2 }+{ \left( 9-5 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }-4a+20 }$ ,

$BC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 5-a \right) }^{ 2 }+{ \left( 5-5 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }-10a+25 }$ and

$AC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( 5-2 \right) }^{ 2 }+{ \left( 5-9 \right) }^{ 2 } }$ units

$=\sqrt { 25 }$ units.

$\therefore$ by (i), we get

${ a }^{ 2 }-4a+20{ +a }^{ 2 }-10a+25=25$

$\Rightarrow { a }^{ 2 }-7a+10=0$

$\Rightarrow a=5,2$

We reject

$a=5$ , since

$B$ and

$C$ will coincide in that case and

$\Delta ABC$ will collapse.

So,

$a=2. i.e. AB=\sqrt { { a }^{ 2 }-4a+20 } =\sqrt { 4-8+20 } 3$ units

$=43$ units

$BC=\sqrt { { a }^{ 2 }-10a+25 } =\sqrt { 4-20+25 } 3$ units

$=3$ units.

Now,

We know that the area of the triangle whose vertices are

$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$ and

$(x_{3},y_{3})$ is

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

Therefore, required area is

$6$ square units.

If (0, 0), (3, 0) and (x, y) are the vertices of an equilateral triangle, then the value of x and y is

0%
$\left ( \frac{3}{2},\frac{3\sqrt{3}}{2} \right )or\left ( ^{3}_{2},\frac{-3\sqrt{3}}{2}\right )$
0%
$\left ( \frac{1}{2},\sqrt{2} \right )or \left ( \frac{1}{2},-\sqrt{2} \right )$
0%
$\left ( \frac{1}{3},1 \right )or \left ( \frac{1}{3}, -1\right )$
0%
none of these

Explanation

$Let\quad us\quad take\quad the\quad equilateral\quad \Delta \quad as\quad AOB\quad with\quad A=\left( x,y \right) ,\quad B=\left( 3,0 \right) ,\quad O=\left( 0,0 \right) .\quad \\ The\quad ordinates\quad of\quad B\quad \& \quad O\quad are=0.\quad So\quad BO\quad lies\quad on\quad the\quad X-axis.\quad Also\\ (AB=BO\quad \& \quad AO=BO).................(i)\quad [since\quad \Delta AOB\quad is\quad equilateral]\\ \therefore \quad The\quad length\quad BO=3-0=3.........(ii)\\ Now\quad AB=\sqrt { { \left( 3-x \right) }^{ 2 }+{ \left( 0-y \right) }^{ 2 } } =\sqrt { { x }^{ 2 }+{ y }^{ 2 }-6x+9 } =3\quad [from\quad (i)\quad ]\\ \Longrightarrow { x }^{ 2 }+{ y }^{ 2 }-6x=0.............(iii)\\ Again\quad AO=\sqrt { { \left( 0-x \right) }^{ 2 }+{ \left( 0-y \right) }^{ 2 } } =3\quad [from\quad (i)\quad ]\\ \Longrightarrow { x }^{ 2 }+{ y }^{ 2 }=9................(iv)\\ \therefore \quad From\quad (i)\quad \& \quad (iv)\quad we\quad have\quad 9-6x=0\Longrightarrow x=\frac { 3 }{ 2 } \\ Substituting\quad x=\frac { 3 }{ 2 } \quad in\quad (iv),\quad y=\pm \frac { 3\sqrt{3} }{ 2 } \\ \therefore \quad \left( x,y \right) =\left( \frac { 3 }{ 2 } ,\frac { 3\sqrt{3} }{ 2 } \right) ,\quad \left( \frac { 3 }{ 2 } ,-\frac { 3\sqrt{3} }{ 2 } \right) \\ Ans-\quad Option\quad A\\$

The area of triangle ABC (in sq. units) is :

0%
$15$ sq. units
0%
$10$ sq. units
0%
$7.5$ sq. units
0%
$2.5$ sq. units

Explanation

The area of a triangle is given as:

Area

$=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right]$

the given points are

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

$C(4,0)$

by Substituting ,

$\text{Area} =\dfrac { 1 }{ 2 } \left[ 1(0-0)+(-1)(0-3)+4(3-0) \right] \\ =\dfrac { 1 }{ 2 } \left[ 3+12 \right] =\dfrac { 15 }{ 2 } =7.5$

$\therefore$ Area of the given triangle ABC is

$7.5$ sq. units

Say true or false.

The area of the triangle formed by the points

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

$(x_3, y_3)$ is the numerical value of the expression

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

0%
True
0%
False

Explanation

Let

$ABC$ be a triangle whose vertices are

${A(x_1,y_1)}$ ,

${B(x_2,y_2)}$ and

${C(x_3,y_3)}$ .
Draw the lines

$AD, BE, CF$ perpendicular to

$x$ -axis.
From the figure,

$ED =$

${x_1} - {x_2}$ ;

$DF =$

${x_3} - {x_1}$ ;

$EF =$

${x_3} - {x_2}$
Area of the triangle

$ABC =$ Area of the trapezium

$ABED +$ Area of the trapezium

$ADFC -$ Area of the trapezium

$BEFC$

$=$

$\dfrac{1}{2} [(BE + AD)ED ]$

$+$

$\dfrac{1}{2} [(AD +CF)DF]$

$-$

$\dfrac{1}{2} [(BE+CF)EF]$

$=$

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

$+$

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

$-$

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

$=$

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

Area of the triangle formed by the points P(-1.5, 3), Q(6, -2) and R(-3, 4) is 0.

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

We know that the area of the triangle whose vertices are

$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$ and

$(x_{3},y_{3})$ is

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

Given

$P(-1.5, 3), Q(6, -2)$ and

$R(-3, 4)$
Therefore, area is given by

$= \frac{1}{2}\times[(-1.5)(-2-4) + 6(4-3)+(-3)(3+2)]$

$= \frac{1}{2}\times(9 +6-15)$

$= 0$

The angle made by the line whose intercepts are 3 and

$3\sqrt 3$ on X and Y axis respectively with X-axis is .....

0%
$30^0$
0%
$120^0$
0%
$60^0$
0%
$90^0$

Explanation

Any straight line which makes an angle

$\alpha$ with the positive direction of

$X-$ axis will have their slope as

$\tan \alpha=-\dfrac{b}{a}$ , where

$a$ and

$b$ are the intercepts w.r.t.

$X$ and

$Y$ axes

Here,

$a=3, b=3\sqrt{3}$

$\implies \tan \alpha =-\dfrac{3\sqrt{3}}{3}=-\sqrt{3}$

$\implies \alpha=\tan^{-1}(-\sqrt{3})$

$\therefore \alpha = 120^{o}$

Hence, option B is correct.

In the given figure if

$O, P, Q$ and

$R$ are the vertices of a square and the co-ordinates of

$P$ are

$(K, p)$ , the values of

$k$ and

$p$ are

0%
$k = -4$ and $p = 0$
0%
$k = 0$ and $p = - 4$
0%
$k = -2$ and $p = 0$
0%
$k = 0$ and $p = -2$

Explanation

It is evident from the figure that

$POQR$ is a square.

Its side

$OR$ lies on the

$x$ -axis.

$\therefore$ the length of

$OR$ is the difference of abscissa between

$O \& R$ .

Now,

$O=\left( 0,0 \right)$ and

$R=\left( 4,0 \right)$

$\therefore OR=4-0=4$ units.

Again

$\Box POQR$ is a square.

$\therefore OR=OP=4$ units.

But

$OP$ lies on the

$y$ -axis to the negative side of it.

$\therefore$ ordinate of

$P$ is

$0-4=4=p$ and

$k=0$

$\therefore P=\left( 0,-4 \right)$

A circle that has its center at the origin and passes through

$(-8, -6)$ will also pass through the point

0%
$(1, 10)$
0%
$(4, 7)$
0%
$(7, 7)$
0%
$(9,\sqrt{19})$

Explanation

Here, it is given that the circle with center at origin passes through the point

$(-8, -6).$
The distance from origin

$(0, 0)$ to point

$(-8, -6)$ will be the radius of the given circle.
So, by distance formula,
Radius of circle

$= \sqrt{(-8-0)^2+(-6-0)^2}$

$= \sqrt{8^2+6^2}$

$= \sqrt{64+34} = \sqrt{100} = 10$ .
Now, any point that lies on the circle will be at the same distance as that of radius.
Hence, check the options whether any point has the same distance from the origin as radius of the circle.

Option (D) gives

$\sqrt{(9-0)^2+(\sqrt{19}-0)^2} = \sqrt{81+19}$

$\sqrt{100} = 10$ .
Hence, the distance of point

$(9, 19)$ from the origin is same as that of radius equal to

$10$ .
The circle given passes through the point

$(9, \sqrt{19})$ .

If the vertices of a triangle are

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

$(1, - 3)$ , then the triangle is

0%
scalene
0%
equilateral
0%
isosceles
0%
right triangle

Explanation

Let the vertices of the triangle be

$A(-2,0)$ ,

$B(2,3)$ and

$C(1,-3)$ .

We compute the length of the sides as follows:

$AB=\sqrt { { (2-(-2)) }^{ 2 }+{ (3-0) }^{ 2 } } =\sqrt { 4^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 16+9 } =5$

$BC=\sqrt { { (1-2) }^{ 2 }+{ (-3-3) }^{ 2 } } =\sqrt { (-1)^{ 2 }+{ (-6) }^{ 2 } } =\sqrt { 1+36 } =\sqrt { 37 }$

$AC=\sqrt { { (1-(-2)) }^{ 2 }+{ (-3-0) }^{ 2 } } =\sqrt { 3^{ 2 }+{ (-3) }^{ 2 } } =\sqrt { 9+9 } =\sqrt { 18 }$

Since none of the lengths of the sides are equal, therefore, it is a scalene triangle.

Perimeter of the triangle whose vertices are

$(3,2) (7,2) (7,5)$ is

0%
$14$
0%
$12$
0%
$10$
0%
$8$

Explanation

Let the given vertices are

$A(3,2),B(7,2),C(7,5)$

Distance formula for two points is

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

Using the formula,

$AB=\sqrt{(7-3)^2+(2-2)^2}$

$AB=\sqrt{4^2+0}$

$AB=\sqrt{4}$

$AN=4$ units

Similarly,

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

$BC=\sqrt{0^2+3^2}$

$BC=\sqrt{3^2}$

$BC=3$ units

$CQ=\sqrt{(3-7)^2+(2-5)^2}$

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

$=\sqrt{16+9}$

$=\sqrt{25}$

$=5$ units

The perimeter of the triangle

$=AB+BC+CA$

$=4+3+5=12$ units

The distance or origin from the point P(3,2) is :

0%
$\sqrt 2$
0%
$\sqrt {15}$
0%
$\sqrt {13}$
0%
$\sqrt {11}$

Explanation

$distance= \sqrt{A^{2}+B^{2}}$

$distance = \sqrt {(0-3)^2+(0+2)^2}$

$=\sqrt {9+4}=\sqrt {13}$ .

In the given figure, which of the following points lies within the circle?

0%
$(3.5, 9.5)$
0%
$(-7, 7)$
0%
$(-10, 1)$
0%
$(-10, -1)$

Explanation

The centre of the circle is at

$\left( 0,0 \right)$ .

$\therefore$ Its equation is

${ x }^{ 2 }+{ y }^{ 2 }={ r }^{ 2 }$ ......(i)

When

$r=$ radious of the circle.

It passas through

$\left( 6,8 \right)$ .

$\therefore$ putting

$x=6$ and

$y=8$ in (i), we get

$6^{ 2 }+{ 8 }^{ 2 }={ r }^{ 2 }\Longrightarrow r=10$ units.

If the distance of any point is less than the length of the radius

$r$ , then it will be within the circle.

Now, we investigate each option which has a point at a distance

$=d$ less than the length of

$r$ .

Option A:

$\longrightarrow { x }_{ 1 }=3.5\ , \ { y }_{ 1 }=9.5$

${ \therefore d }^{ 2 }={ { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }={ 3.5 }^{ 2 }+{ 9.5 }^{ 2 }=102.5\Longrightarrow d>10$

Option B:

$\longrightarrow { x }_{ 1 }=-7\ ,\ { y }_{ 1 }=7$

${ \therefore d }^{ 2 }={ { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }=({ -7) }^{ 2 }+{ 7 }^{ 2 }=93.50\Longrightarrow d<10$

Option C:

$\longrightarrow { x }_{ 1 }=-10\ , \ { y }_{ 1 }=1$

${ \therefore d }^{ 2 }={ { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }=({ -10) }^{ 2 }+{ 1 }^{ 2 }=101.1\Longrightarrow d>10$

Option D:

$\longrightarrow { x }_{ 1 }=-10\ , \ { y }_{ 1 }=-1\\ { \therefore d }^{ 2 }={ { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }=({ -10) }^{ 2 }+(-{ 1) }^{ 2} = 101.1 \Longrightarrow d>10$

$\therefore$ Only

$\left( -7,7 \right)$ lies within the circle.

If A

$(-3, 0)$ and C

$(5, 2)$ are the end points of diagonal

$AC$ of rectangle

$ABCD$ , with

$B$ on the x- axis, the perimeter of the rectangle

$ABCD$ is

0%
$20$
0%
$24$
0%
$28$
0%
$30$

Explanation

Both A & B i.e the side AB lies on the X-axis.

Again

$BC\bot AB$ ...(since

$\Box ABCD$ is a rectangle)

$\Rightarrow$ Abscissa of

$B=$ abscissa of

$C$ .

$\therefore B=\left( x,0 \right) =\left( 5,0 \right)$ ....[since

$C=\left( 5,2 \right)$ ]

$\therefore$ Length

$AB=$ difference beteen abscissae of A & B

$=[\left( 5-(-3 \right) ]=8$ units.

Again length

$BC=$ difference beteen ordinates of B &

$C=\left( 2-0 \right) =2$ units.

$\therefore$ The perimeter

$P=2\left( AB+BC \right) =2\left( 8+2 \right) =20$ sq. units.

On the Cartesian plane. PQR is an isosceles triangle.
The perimeter of

$\Delta PQR$ is

0%
16 units
0%
18 units
0%
20 units
0%
22 units

Explanation

$PQ=\sqrt{3^{2}+4^{2}}=5 units =PR$

$QR=8-2=6$
Perimeter of

$\Delta PQR=PQ+QR+PR$

$=5+6+5$

$=16 units$

If the point P(x, y) is equidistant from the points A

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

$(a-b, a+b)$ , then which of the following condition is true?

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

Explanation

We have

$PA=PB$ , i.e.,

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

$\Rightarrow \left [ x-(a+b) \right ]^{2}+\left [ y-(b-a) \right ]^{2}$

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

$\Rightarrow \left [ (x-a)-b \right ]^{2}+\left [ (y-b)+a \right ]^{2}$

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

$\Rightarrow \left [ (x-a)+b \right ]^{2}-\left [ (x-a)-b \right ]^{2}$

$=\left [ (y-b)+a \right ]^{2}-\left [ (y-b)-a \right ]^{2}$

$\Rightarrow 4b(x-a)=4a(y-b)$

$\Rightarrow bx=ay$ ........(1)
Therefore, (B) is correct. Also, P(a, b) satisfies the condition (1), so that P can be (a, b) and hence (D) is
also correct.

The distance between which two points is 2 units ?

0%
$(-2, -3)$ and $(-2, -4)$
0%
$(0, 4)$ and $(6, 0)$
0%
$(7, 2)$ and $(6, 2)$
0%
$(4, -3)$ and $(2, -3)$

Explanation

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

$=\sqrt{2^{2}+0^{2}}$

$=2 units$

The coordinates of the point on the

$\displaystyle x-$ axis which is equidistant from the points

$\displaystyle(-3,4)$ and

$\displaystyle(2,5)$ are :

0%
$\displaystyle(20,0)$
0%
$\displaystyle(-23,0)$
0%
$\left(\dfrac{4}{5},0\right)$
0%
none of these

Explanation

Using distance formula, we get because point

$(p,0)$ is at equidistant from

$(-3,4)$ and

$(2,5)$

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

$p=\dfrac{2}{5}$
Hence, option 'D' is correct.

A(3, -4), B(-5, 6), then the distance between these two points is

0%
$\sqrt{41}$
0%
$3\sqrt{41}$
0%
$4 \sqrt{41}$
0%
$2 \sqrt{41}$

Explanation

The given points are

$A=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 3,-4 \right) and B=\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -5,6 \right)$

The distance

$d$ between

$A$ and

$B$ is

$d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 3+5 \right) }^{ 2 }+{ \left( -4-6 \right) }^{ 2 } }$ units

$=2\sqrt { 41 }$ units

The diagram shows a rectangle on a Cartesian plane. M is the midpoint of PQ.
The coordinate of R are:

0%
(9, 6)
0%
(8, 6)
0%
(9, 1)
0%
(10, 6)

Explanation

y-component of R is 6
x-component of R

$=x$ component of Q

$i.e., \dfrac{x+1}{2}=5$

$x=9$

$\therefore R=(9, 6)$

P is the point

$(-5, 3)$ and Q is the point

$(-5, m)$ . If the length of the straight line PQ is 8 units, the the possible value of "m" is:

0%
$-5$ and 5
0%
$-5$ or 11
0%
$-5$ or $-11$
0%
5 or 11

Explanation

Since x-component of P and Q are same i.e., -5
y-component of Q should be 8 units above or below.
i.e., 11 or

$-5$

What is the length of the line segment AB?

0%
8 units
0%
1 units
0%
13 units
0%
15 units

Explanation

We have,

$A=(-1,3),B=(4,-9)$

We know that the distance between two points

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

Therefore,

$d=\sqrt{(4-(-1))^2+(-9-3)^2}$

$d=\sqrt{(4+1)^2+(-12)^2}$

$d=\sqrt{(5)^2+(-12)^2}$

$d=\sqrt{25+144}$

$d=\sqrt{169}$

$d=13\ units$

Hence, this is the answer.

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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