MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 4

In the diagram MN, is a straight line.
The distance between M and N is:

0%
6 units
0%
8 units
0%
9 units
0%
10 units

Explanation

$M(3, 2); N(-3, -6)$

$\therefore MN=\sqrt{36+64}$

$=10 units$

On the Cartesian plane, Q is the midpoint of the straight line PR.
Find the value of h.

0%
4
0%
5
0%
7
0%
8

Explanation

(C)

$\dfrac{h-3}{2}=2$

$h=4+3=7$

$P(x_{1}, y_{1})$ and

$Q(x_{2}, y_{2})$ are points in the plane such that

$PQ$ subtends a right angle at the origin

$O$ , then

0%
$P(1, 3), Q(-3, 1)$
0%
$P(3, 1), Q(1,-3)$
0%
$P(2, 5), Q(5, -2)$
0%
$P(1, 1), Q(-1, 1)$

Explanation

$PQ$ subtends a right angle at the origin

$O,$ then.

$\Rightarrow \displaystyle { \left( OP \right) }^{ 2 }+{ \left( OQ \right) }^{ 2 }={ \left( PQ \right) }^{ 2 }$

$\Rightarrow (x_1^2+y_1^2)+(x_2^2+y_2^2)=(x_1-x_2)^2+(y_1-y_2)^2$

$\displaystyle \Rightarrow \left( { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 } \right) =0$

Then from aptions

$\displaystyle \left( A \right) \quad 1.\left( -3 \right) +\left( 3 \right) .\left( 1 \right) =-3+3=0$

$\displaystyle \left( B \right) \quad 3.1+1.\left( -3 \right) =3-3=0$

$\displaystyle \left( C \right) \quad 2.5+\left( 5 \right) .\left( -2 \right) =10-10=0$

$\displaystyle \left( D \right) \quad 1.\left( -1 \right) +\left( 1 \right) .\left( 1 \right) =1+1=0$

Hence All the options are true.

The co-ordinate of a point R on a line is 8, The point S is on the same line which is to the left of R and a distance of 7 units from R. Find the co-ordinate of S. If P is the midpoint of seg RS,find the co-ordinate of point P.

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Co-ordinate of S is 0. P $\leftrightarrow$ 3
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Co-ordinate of S is 1. P $\leftrightarrow$ 3.5
0%
Co-ordinate of S is 2. P $\leftrightarrow$ 34
0%
Co-ordinate of S is 4. P $\leftrightarrow$ 5

Explanation

The co-ordinate of a point R on a line is 8, The point S is on the same line which is to the left of R and a distance of 7 units from R.
Co-ordinate of point

$S = 8 -7 = 1$
P is the midpoint of segment RS
Co-ordinate of point P = midpoint of Co-ordinate of point S and Co-ordinate of point R

$= \dfrac {8-1}{2}$

$= 3.5$

The points

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

$(0, -1)$ are the vertices of a ________.

0%
parallelogram.
0%
scalene quadrilateral
0%
isosceles quadrilateral
0%
concave quadrilateral

Explanation

If the points are

$P(x_{1},y_{1})$ and

$Q (x_{2}, y_{2})$ , the slope of the line joining PQ is

$\dfrac{y_{2} - y_{2}}{x_{2} - x_{1}}$

For points,

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

$D (0, -1)$ ,
Slope of

$AB = \dfrac{3 - 1}{0 + 2} = 1$

Slope of

$BC = \dfrac{1 - 3}{2 - 0} = -1$

Slope of

$CD = \dfrac{-1 - 1}{0 - 2} = 1$
Slope of

$DA = \dfrac{-1 - 1}{0 + 2} = -1$

Clearly slope of

$AB =$ Slope of

$CD$ , hence,

$AB \parallel CD$ .
Clearly slope of

$BC =$ Slope of

$DA$ , hence,

$BC \parallel DA$ .

Since, the set of opposite sides are parallel to each other.

Thus,

$\Box ABCD$ are the vertices of a parallelogram.

For the equation given below, find the slope and the y-intercept:

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

0%
$\displaystyle \frac{1}{3} \ and \ \frac{5}{3}$
0%
$\displaystyle -\frac{1}{3} \ and \ -\frac{5}{3}$
0%
$\displaystyle -{3} \ and \ \frac{3}{5}$
0%
$\displaystyle {3} \ and \ -\frac{5}{3}$

Explanation

The equation of any straight line can be written as

$y = mx + c$ ,

where

$m$ is its slope and

$c$ is its y - intercept.

$x + 3y + 5 = 0$ can be written as

$3y = -x - 5$ or

$y = \cfrac {-1}{3}x - \cfrac {5}{3}$

Comparing this equation with the standard form of the equation, we get:

$m = - \cfrac {1}{3} , c = - \cfrac {5}{3}$

Hence, slope of

$x + 3y + 5 = 0$ is

$- \cfrac {1}{3}$ and y - intercept is

$- \cfrac {5}{3}$

Two vertices of an equilateral triangle are

$(-1,0), (1, 0)$ . Third vertex can be

0%
$\left ( \sqrt{3}, 0 \right )$
0%
$\left ( 0, \sqrt{3} \right )$
0%
$\left ( 0, -\sqrt{3} \right )$
0%
$\left ( 0, 2\sqrt{3} \right )$

Explanation

The distance between the two given vertices of the equilateral traingle is

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

Hence, length of each side of equilateral triangle is

$2$ units.

$\sin 60 = \dfrac {\text{perpendicular}}{\text{hyp}}$

$\dfrac {\sqrt 3}{2} = \dfrac {\text{perpendicular}}{2}$

$\therefore \text {perpendicular} = \sqrt 3$ units.

Hence, the coordinates of the

$3^{rd}$ vertex can be

$(0, \sqrt 3)$ or

$(0, \sqrt {-3})$ since it lies on the

$Y$ -axis.

$ABCD$ is a rectangle with

$A(-1, 2)$ ,

$B ( 3, 7 )$ and

$AB : BC= 4 : 3$ . If

$P$ is the centre of the rectangle then the distance of

$P$ from each corner is equal to

0%
$\displaystyle \dfrac{\sqrt{41}}2$
0%
$\displaystyle \dfrac{3\sqrt{41}}4$
0%
$\displaystyle \dfrac{2\sqrt{41}}3$
0%
$\displaystyle \dfrac{5\sqrt{41}}8$

Explanation

Since

$ABCD$ is a rectangle,

$\left ( AC \right )^{2}=\left ( AB \right )^{2}+\left ( BC \right )^{2}$

$=\left ( AB \right )^{2}\left [ 1+\dfrac {9}{16} \right ]=\dfrac{25}{16}\left ( AB \right )^{2}$ . ...

$\left[\because BC = \dfrac 34 AB\right]$

$\therefore AC = \dfrac 54AB$

Now,

$PA=\dfrac 12 AC=\dfrac 58 AB$

By distance formula, we have

$AB=\sqrt{\left ( -1-3 \right )^{2}+\left ( 2-7 \right )^{2}}=\sqrt {41}$ .

$\Rightarrow AC = \dfrac 54 \sqrt {41}$

$\Rightarrow PA = \dfrac 58 \sqrt {41}$

$\displaystyle \alpha$ is a root of the equation:

$x^{2}-5x + 6 =0$ and

$\displaystyle \beta$ is a root of the equation

$x^{2}-x -30 = 0,$ then coordinates

$(\alpha,\beta)$ of the point P farthest from the origin are

0%
$(2,6)$
0%
$(2,3)$
0%
$(6,-5)$
0%
$(3,6)$

Explanation

${x}^{2}-5x+6=0$

$(x-3)(x-2)=0$

$\therefore \alpha$ can be

$3$ or

$2.$

${x}^{2}-x-30=0$

$(x-6)(x+5)=0$

$\therefore \beta$ can be

$6$ or

$-5.$
So points which is farthest from origin is

$(3,6).$

In the figure OAB is an equilateral triangle. The co-ordinate of vertex B is

0%
$(a, a)$
0%
$(-a, -a)$
0%
$(a, \sqrt{3}a)$
0%
$(a, -\sqrt{3}a)$

Explanation

In figure,
OAB is an equilateral triangle of side

$2a$ .

$\therefore OA = AB = OB = 2a$
Now, from the point B, draw BM perpendicular on OA.

$\therefore OM = MA = a$
Therefore, from right

$\triangle OMB$ ,

$OB^2 = OM^2+MB^2$

$(2a)^2 = a^2+MB^2$

$MB^2 = 3a^2$

$\therefore MB = \sqrt{3}a$
Since

$OM = a$ and

$MB = \sqrt 3$

Hence, co-ordinates of vertex B are

$(a, \sqrt 3a)$

Find a relation between x and y such that the point

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

$(3, 6)$ and

$(-3, 4)$ .

0%
$3x+y-5 =0$
0%
$3x-y-5 =0$
0%
$3x+y+5 =0$
0%
$3x-y+5 =0$

Explanation

Distance between $P (x,y)$ and $B (-3,4) = PB = \sqrt { \left( -3-x \right) ^{ 2 }+\left( 4 - y \right) ^{ 2 } } = \sqrt { 9 + { x }^{ 2 } + 6x + 16 + { y }^{ 2 } - 8y } = \sqrt { { x }^{ 2 } + 6x + { y }^{ 2 } - 8y + 25 }$

Given, PA $=$ PB $\sqrt { { x }^{ 2 } - 6x + { y }^{ 2 } - 12y + 45 } = \sqrt { { x }^{ 2 } + 6x + { y }^{ 2 } - 8y + 25 }$

$=> { x }^{ 2 } - 6x + { y }^{ 2 } - 12y + 45 = { x }^{ 2 } + 6x + { y }^{ 2 } - 8y + 25$

$12x + 4y - 20 = 0$
$3x + y - 5 = 0$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

Reason is true.
Because if

$ABC$ is a given triangle and

$D, E, F$ are the mid-points of the sides

$BC, CA$ and

$AB$ respectively, then
Area of

$\Delta AEF$ .

$=\dfrac{1}{2}\times \dfrac{1}{2}BC\times \dfrac{1}{2}\:l$ (l being the altitude from A)

$=\dfrac{1}{4}\times \dfrac{1}{2}BC\times l=\dfrac{1}{4}$ area of the triangle

$ABC.$
Similarly for triangles

$BDF$ and

$CDE,$
So that area of

$\Delta DEF=\dfrac{1}{4}$ area of

$\Delta ABC$

$\displaystyle =\dfrac{1}{4}\times \dfrac{1}{2}\begin{vmatrix} 2 & 2 & 1 \\ 6 & -1 & 1 \\ 7 & 3 & 1 \end{vmatrix}=\frac{19}{8}$
and thus assertion is also true.

What point on y-axis is equidistant from the points

$(3,1)$ and

$(1,5)$ ?

0%

$P(1,3)$
0%
$P(0,2)$
0%
$P(1,2)$
0%
$P(0,3)$

Explanation

Let the point on the y-axis be, $(0,y)$

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

As the point, $(0,y)$ is equidistant from the two points, distance

between $(0,y) ; (3,1)$ and $(0,y) ; (1,5)$ are equal.

$\sqrt { { y}^{ 2 } - 2y + 10 } = \sqrt { { y }^{ 2 } - 10y + 26 }$

$\Rightarrow { y }^{ 2 } - 2y + 10 = { y }^{ 2 } - 10y + 26$

$8y = 16$

$y = 2$

Thus, the point is $(0,2)$

Find the area of the triangle whose vertices are:
(i)

$(2,3)$ ,

$(-1,0)$ ,

$(2,-4)$
(ii)

$(-5,-1)$ ,

$(3,-5)$ ,

$(5,2)$

0%

(i) 32 sq. units
0%
(ii) 32 sq. units
0%
(ii) 10.5 sq. units
0%
(i) 10.5 sq. units

Explanation

Area of a triangle with vertices

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

$({ x }_{ 2 },{ y }_{ 2 })$ and

$({ x }_{ 3 },{ y }_{ 3 })$ is

$\left| \cfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

1. Area of triangle with vertices,

$(2,3), (1,0), (2,4) = \left| \cfrac { 2(0+4) -1(-4-3) + 2(3-0) }{ 2 } \right| = \left| \cfrac { 8+ 7 +6 }{ 2 } \right| = \cfrac {21}{2} sq units$

2.Area of triangle with vertices,

$(-5,-1), (3,-5), (5,2) = \left| \cfrac { -5(-5-2) +3(2+1) + 5(-1+5) }{ 2 } \right| = \left|\cfrac { 35 + 9 +20 }{ 2 } \right| = \cfrac {64}{2} = 32$ sq units

Consider the points

$A( a, b + c)$ ,

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

$C(c, a +b)$ be the vertices of

$\bigtriangleup$ ABC. The area of

$\bigtriangleup$ ABC is:

0%
$2(a^2 + b^2 +c^2)$
0%
$a^2 + b^2 +c^2$
0%
$2(ab + bc + ca)$
0%
None of these

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

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

Since, vertices are

$A(a, b + c),~ B(b, c + a),$ and

$C(c, a + b)$

$\therefore$ Area

$\bigtriangleup ABC = \dfrac{1}{2}|a(c + a) -b(b + c) + b(a + b) - c(c + a) +c(b + c) -a(a + b)|=0$ .

Hence option 'D' is correct.

If the line

$\sqrt{5}x=y$ meets the lines

$x=1, x=2, ..., x=n,$ at points

$A_{1}$ ,

$A_{2}$ , ...,

$A_{n}$ respectively then

$\left ( OA_{1} \right )^{2}+\left ( OA_{2} \right )^{2}+...+\left ( OA_{n} \right )^{2}$ is equal to (O is the origin)

0%
$3n^{2}+3n$
0%
$2n^{3}+3n^{2}+n$
0%
$3n^{3}+3n^{2}+2$
0%
$\left ( \dfrac32 \right )\left ( n^{4}+2n^{3}+n^{2} \right )$

Explanation

Let

$A_{k}(k,\sqrt{5}k)$ be the general point where the given line meets

$x=k$
The distance of this point from the origin is

$OA_{k} =\sqrt{(k-0)^{2}+(\sqrt{5}k-0)^{2}} = \sqrt{6}k$ .
Then,

$OA_{1}^{2} + OA_{2}^{2}+ ................... OA_{n}^{2} = 6(1^{2}+2^{2}+........n^{2})$
=

$2n^{3}+3n^{2}+n$

Find the area of the triangle whose vertices are

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

$(2,3)$ .

0%
$8$ sq.unit
0%
$7$ sq.unit
0%
$6$ sq.unit
0%
$5$ sq.unit

Explanation

Area of a triangle with vertices

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

$({ x }_{ 2 },y_2)$ and

$({ x }_{ 3 },{ y }_{ 3 })$

$= \left |\dfrac{ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3 (y_1 - y_2) } {2} \right |$

Hence, substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (3,2)$ ;

$({ x }_{ 2 },{ y }_{ 2 }) = (-2,-3)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (2,3)$

in the area formula, we get

Area =

$\left | \dfrac {3(-3-3) +(-2)(3-2) + 2 (2-(-3)) }{2}\right | = 5$ sq. unit

Find the value of k, if the point

$(2,3)$ is equidistant from the points

$A(k,1)$ and

$B(7,k)$

0%
$k = 17$
0%
$k = 10$
0%
$k = 13$
0%
$k = 16$

Explanation

$\sqrt { \left( k - 2 \right) ^{ 2 }+\left( 1 - 3 \right) ^{ 2 } } = \sqrt { \left( 7 - 2 \right) ^{ 2 }+\left( k - 3 \right) ^{ 2 } }$

$=> \left( k- 2 \right) ^{ 2 }+ 4 = 25 +\left( k - 3 \right) ^{ 2 }$

On expanding the squares and simplifying, we get $k = 13$

Find the coordinates of the point equidistant from three given points

$A(5,1)$ ,

$B(-3, -7)$ and

$C(7, -1)$ .

0%
$(-2, -4)$
0%
$(2, -4)$
0%
$(2, 4)$
0%
$(-2, 4)$

Explanation

Let the required point be

$P(p,q).$

Then the distances

$PA=PB=PC=d$ as

$A(5,1), B(-3,-7)$ &

$C(7,-1)$ are equidistant from

$P.$

Applying the distance formula,

$P(p,q)\equiv(x_1,y_1)$ and

$A(5,1)\equiv(x_2,y_2)$

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

${ d }_{ PA }=\sqrt { { \left( p-5 \right) }^{ 2 }+{ \left( q-1 \right) }^{ 2 } } ,$

$P(p,q)\equiv(x_1,y_1)$ and

$B(-3,-7)\equiv(x_2,y_2)$

${ d }_{ PB }=\sqrt { { \left( p+3 \right) }^{ 2 }+{ \left( q+7 \right) }^{ 2 } }$ and

$P(p,q)\equiv(x_1,y_1)$ and

$C(7,-1)\equiv(x_2,y_2)$

${ d }_{ Pc }=\sqrt { { \left( p-7 \right) }^{ 2 }+{ \left( q+1 \right) }^{ 2 } }$

$\therefore \sqrt { { \left( p-5 \right) }^{ 2 }+{ \left( q-1 \right) }^{ 2 } } =\sqrt { { \left( p+3 \right) }^{ 2 }+{ \left( q+7 \right) }^{ 2 } }$

$\Longrightarrow -16p-16q=32$

$\Longrightarrow p+q=-2$ ........(i)

$\sqrt { { \left( p+3 \right) }^{ 2 }+{ \left( q+7 \right) }^{ 2 } } =\sqrt { { \left( p-7 \right) }^{ 2 }+{ \left( q+1 \right) }^{ 2 } }$

$\Longrightarrow 20p+12q=-8$

$\Longrightarrow 5p+3q=-2$ .........(ii).

Multiplying (i) by 5 & subtracting from (ii),

$-2q=8$

$\Longrightarrow q=-4.$

Substituting

$q=-4$ in (i),

$p=-2-q=-2+4=2.$

$\therefore P(p,q)=P(2,-4).$

Find the value of k for which the distance between the points

$A(3k,4)$ and

$B(2,k)$ is

$5\sqrt { 2 }$ units.

0%
$k = -1$
0%
$k =3$
0%
$k =-3$
0%
$k =1$

Explanation

Distance

between two points

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula

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

$(3k,4)$ and

$(2,k) = \sqrt { \left( 3k-2 \right) ^{ 2 }+\left( 4 - k \right) ^{ 2 } } = \sqrt { 9{k}^{2} + 4 - 12k + 16 + {k}^{2} - 8k } = \sqrt { 10{ k}^{ 2 } - 20k + 20 }$

Given, distance between the points

$= 5\sqrt { 2 }$

$=> \sqrt { 10{ k }^{ 2 } - 20k + 20 } = 5\sqrt { 2 }$

$=> 10{ k }^{ 2 } - 20k + 20 = 25\times 2 = 50$

$=> { k }^{ 2 } - 2k + 2= 5$

$=> { k }^{ 2 } - 2k - 3 = 0$

$=> { k }^{ 2 } - 3k + k - 3 = 0$

$=> k(k - 3) + 1( k - 3) = 0$

$=> (k+ 1)(k - 3)= 0$

$=> k = -1,3$

If two vertices of an equilateral triangle are

$(0,0)$ and

$(3, \sqrt { 3 })$ , find the third vertex of the triangle

0%
$(0, 2\sqrt { 3 })$
0%
$(0, -\sqrt { 3})$
0%
$(3, -\sqrt { 3})$
0%
$(3, 2\sqrt { 3 })$

Explanation

Let (x,y) be the third vertex of the triangle.

Distance between (0,0) and

$(3,\sqrt3)$ is

$d=\sqrt{3^2+\sqrt3^2}=\sqrt{12}=2\sqrt3$

Since the triangle is equilateral, therefore all the sides are equal.

$\Rightarrow (x-0)^2+(y-0)^2=12$

$\Rightarrow x^2+y^2=12$ ...(i)

Again,

$(x-3)^2+(y-\sqrt3)^2=12$

$\Rightarrow x^2-6x+9+y^2-2\sqrt3y+3=12$

$\Rightarrow (x^2+y^2)-6x-2\sqrt3y=0$

$\Rightarrow (12)-6x-2\sqrt3y=0$ (from (i))

$\Rightarrow 3x+\sqrt3y=6$

$\Rightarrow x=\cfrac{6-\sqrt3y}{3}$

on putting this value in (i), we get

$(\cfrac{6-\sqrt3y}{3})^2+y^2=12$ ...(ii)

$\Rightarrow (\cfrac{36+3y^2-12\sqrt3y}{9})+y^2=12$

$\Rightarrow 12y^2-12\sqrt3y=72$

$\Rightarrow y^2-\sqrt3y-6=0$

$\Rightarrow y^2-2\sqrt3y+\sqrt3y-6=0$

$\Rightarrow y(y-2\sqrt3)+\sqrt3(y-2\sqrt3)=0$

$\Rightarrow y=2\sqrt3,-\sqrt3$

form (ii)

when,

$y=-\sqrt3;x=3$

$y=2\sqrt3;x=0$

Therefore, points are

$(0,2\sqrt3)$ or

$(3,-\sqrt3)$

A point P lies on the x-axis and has abscissa

$5$ and a point Q lies on y-axis and has ordinate

$-12$ . Find the distance PQ

0%
$13$ units
0%
$8$ units
0%
$15$ units
0%
$11$ units

Explanation

Point

$P = (5,0)$ ;

$Q = (0,-12)$

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

Distance between the points P and Q $= \sqrt { \left( 0 - 5 \right) ^{ 2 }+\left( -12 \right) ^{ 2 } } = \sqrt { 22 + 144 } = \sqrt { 169 } = 13$

Find a point on the y-axis which is equidistant from the points

$(-3,4)$ and

$(2,3)$ .

0%
$(0,-6)$
0%
$(0,6)$
0%
$(0,3)$
0%
$(0,-3)$

Explanation

Let the point on the x-axis be $(0,y)$

Distance between $(0,y)$ and $(-3,4) = \sqrt { \left( -3-0 \right) ^{ 2 }+\left( 4 - y \right) ^{ 2 } } = \sqrt { 9 + { 4}^{ 2 }+ { y }^{ 2 } - 8y } = \sqrt { { y }^{ 2 } - 8y + 25 }$

Distance between $(0,y)$ and $(2,3) = \sqrt { \left( 2-0 \right) ^{ 2 }+\left( 3 - y \right) ^{ 2 } } = \sqrt { 4 + { 3}^{ 2 }+ { y }^{ 2 } - 6y } = \sqrt { { y }^{ 2 } - 6y + 13 }$

As the point $(0,y)$ is equidistant from the two points, both the distances

calculated are equal.
$\sqrt { { y }^{ 2 } - 8y + 25 } = \sqrt { { y }^{ 2 } - 6y + 13 }$
$=> { y }^{ 2 } - 8y + 25 = { y }^{ 2 } - 6y + 13$

$12 = 2y$

$y = 6$

Thus, the point is $(0,6)$

Find a relation between x and y such that the point

$(x,y)$ is equidistant from

$(7,1)$ and

$(3,5)$

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

Explanation

${\textbf{Step -1: Write the assumption value}}{\textbf{.}}$

${\text{Let }}P\left( {x,y} \right){\text{ be the equation from the points }}A\left( {7,1} \right){\text{ and }}B\left( {3,5} \right).$

${\textbf{Step -2: Solve according to the formula of equidistant}}{\textbf{.}}$

$AP = BP$

$\Rightarrow A{P^2} = B{P^2}$

$\Rightarrow {\left( {x - 7} \right)^2} + {\left( {y - 1} \right)^2} = {\left( {x - 3} \right)^2} + {\left( {y - 5} \right)^2}$

$\Rightarrow{x^2} - 14x + 49 + {y^2} - 2y + 1 = {x^2} - 6x + 9 + {y^2} - 10y + 25$

$\Rightarrow x - y = 2$

${\textbf{Hence, the relation between }}{\mathbf{x}}{\textbf{ and }}{\mathbf{y}}{\textbf{ is (A) }}{\mathbf{x - y = 2.}}$

In figure, find the coordinates of the centre of the circle which is drawn through the points A, B and O.

0%
$\begin{pmatrix} \dfrac {25 }{ 14 },\dfrac { 25 }{ 14 }\end{pmatrix}$
0%
$\begin{pmatrix} \dfrac {15 }{ 14 },\dfrac { 15 }{ 14 }\end{pmatrix}$
0%
$\begin{pmatrix} \dfrac {15 }{ 14 },\dfrac { 25 }{ 14 }\end{pmatrix}$
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$\begin{pmatrix} \dfrac {25 }{ 14 },\dfrac { 15 }{ 14 }\end{pmatrix}$

Explanation

Given-

Points

$A({ x }_{ 1 },{ y }_{ 1 })=A(-1,2), O({ x }_{ 2 },{ y }_{ 2 })=O(0,0)$ &

$B({ x }_{ 3 },{ y }_{ 3 })=B(3,1)$ are points on a circle.

To find out-

The coordinates of the centre P

Solution-

Let the centre be

$P(x,y).$

Then,

$PA=PO=PB$ since they are radii of the same circle.

We shall use distance formula

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

So,

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

$PO=\sqrt { { \left( { x }_{ 2 }-{ x } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y } \right) }^{ 2 } } =\sqrt { { \left( 0+{ x } \right) }^{ 2 }+{ \left( 0+{ y } \right) }^{ 2 } } =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$

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

Now,

$PA=PO$

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

$\Longrightarrow 2x-4y+5=0$ .........(i)

$PO=PC$

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

$\Longrightarrow -6x-2y+10=0$ .........(ii).

Multiplying (ii) by 3 and adding to (ii),

$14y=25$

$\Longrightarrow y=\cfrac { 25 }{ 14 }$

Substituting

$y=\cfrac { 25 }{ 14 }$ in (i),

$2x-4\times \cfrac { 25 }{ 14 } +5=0$

$\Longrightarrow x=\cfrac { 15 }{ 14 } .$

$\therefore$ The coordinates of the centre are

$\left( \cfrac { 25 }{ 14 } ,\cfrac { 15 }{ 14 } \right) .$

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

$(2,2)$ ,

$(4,4)$ and

$(2,6)$ .

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5
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3
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1
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0

Explanation

Let

$A(2,2), B(4,4)$ and

$C(2,6)$ be the vertices of a given triangle ABC.

Let D, E, and F be the midpoints of AB, BC and CA respectively.

Mid point of two points

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

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

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

Using this formula,the coordinates of D, E, and F are given as

$\displaystyle D \left (\frac{2+4}{2},\frac {2+4}{2} \right ), E\left (\frac {4+2}{2},\frac {4+6}{2}\right )$ and

$F\left (\dfrac {2+2}{2},\dfrac {2+6}{2} \right )$

i.e.,

$D(3,3), E(3,5) and F(2,4)$

Area of a triangle with vertices

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

$({ x }_{ 2 },{ y }_{ 2 })$ and

$({ x }_{ 3 },{ y }_{ 3 })$ is

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

Hence, substituting the points

$({ x }_{ 1 },{ y }_{ 1 }) = (3,3)$ ;

$({ x }_{ 2 },{ y }_{ 2 }) = (3,5)$ and

$({ x }_{ 3 },{ y }_{ 3 }) = (2,4)$

in the area formula, we get

Area of triangle DEF

$= \left| \dfrac { 3(5-4)+(3)(4-3)+2(3-5) }{ 2 } \right| = \left| \dfrac { 3 +3 -4 }{ 2 } \right| = \dfrac {2}{2} = 1 \ sq \ units$

Find the value of

$x$ if the distance between the points

$(2, -11)$ and

$(x, -3)$ is

$10$ units.

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$6$
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$8$
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$9$
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$4$

Explanation

Distance between two points

$= \sqrt {\left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$
Distance between the points

$(2,-11)$ and

$(x,-3) = 10$

$\Rightarrow \sqrt { \left( x- 2 \right) ^{ 2 }+\left( -3 + 11 \right) ^{ 2 } } = 10$

$\Rightarrow \sqrt { { (x - 2) }^{ 2 } + {8}^{2} } = 10$

$\Rightarrow { { (x - 2) }^{ 2 }+ 64 } = 100$

$\Rightarrow { (x - 2) }^{ 2 } = 36$

$\Rightarrow x - 2 = + 6$ or $- 6$

$\Rightarrow x = 8$ or $-4$

Find the area of the right-angled triangle whose vertices are

$(2, -2)$ ,

$(-2, 1)$ and

$(5, 2).$

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$\displaystyle 5\sqrt{2}$ sq. units
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$\displaystyle \dfrac{25}{2}$ sq. units
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$\displaystyle 15\sqrt{2}$ sq. units
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$10$ sq. units

Explanation

Hence, Length of side AB $= \sqrt { \left(-2-2 \right) ^{ 2 }+\left(1 + 2\right) ^{ 2 } } = \sqrt { 16+ 9 } = \sqrt { 25 } = 5$

Length of side BC $= \sqrt {\left(5 + 2\right) ^{ 2 }+\left(2-1\right) ^{ 2 } } = \sqrt { 49 + 1 } = \sqrt { 50}$

Length of side AC $= \sqrt { \left(5-2 \right) ^{ 2 }+\left(2+ 2\right) ^{ 2 } } = \sqrt { 9+16 } = \sqrt { 25 } = 5$

Since,

${(\sqrt { 50 }) }^{2} = { 5 }^{2} + { 5 }^{2}$ ,

$\Rightarrow {BC}^{2} ={AB}^{2} + {AC}^{2}$
Hence, the triangle has a right angle at

$A$ , with

$AB$ and

$AC$ as base and height.
So, area of the triangle

$ABC = \cfrac {1}{2} \times base \times height = \cfrac {1}{2} \times 5 \times 5= \cfrac {25}{2}$ sq units.

The area of the triangle whose vertices are

$A(1,1), B(7, 3)$ and

$C(12, 2)$ is

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$25$ square units
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$8$ square units
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$16$ square units
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$12$ square units

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

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

$\therefore$

Area

$= \dfrac{1}{2} [1(1)+ 7(1)+ 12(-2)] = -8$ , but area can not be negative

Hence, area = 8 square units.

Find the point on the x-axis which is equidistant from the points

$(-2,5)$ and

$(2, -3)$ . Hence find the area of the triangle formed by these points

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$(-2,0);$ $10$ sq. units
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$(-2,5);$ $10$ sq. units
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$(-5,0);$ $10$ sq. units
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$(-5,2);$ $10$ sq. units

Explanation

Distance between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula $\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$
Let the point on the x-axis be $(x,0)$
Distance between $(x,0)$ and $(-2,5) = \sqrt { \left( -2-x \right) ^{ 2 }+\left( 5 - 0 \right) ^{ 2 } } = \sqrt { 4 + { x }^{ 2 } + 4x + 25 } = \sqrt { { x }^{ 2 } + 4x + 29 }$
Distance between $(x,0)$ and $(2,-3) = \sqrt { \left( 2-x \right) ^{ 2 }+\left( -3 - 0 \right) ^{ 2 } } = \sqrt { 4 + { x }^{ 2 } - 4x + 9 } = \sqrt { { x }^{ 2 } - 4x + 13 }$
As the point $(x,0)$ is equidistant from the two points, both the distances calculated are equal.
$\sqrt { { x }^{ 2 } + 4x + 29 } = \sqrt { { x }^{ 2 } - 4x + 13 }$
$=> { x }^{ 2 } + 4x + 29 = { x }^{ 2 } - 4x + 13$
$4x + 4x = 13-29$
$8x = -16$
$x = -2$
Thus, the point is $(-2,0)$

Area of a triangle with vertices $({ x }_{ 1 },{ y }_{ 1 })$ ; $({ x }_{ 2 },{ y }_{ 2 })$ and $({ x }_{ 3 },{ y }_{ 3 })$ is $\left| \dfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$
Hence, substituting the points $({ x }_{ 1 },{ y }_{ 1 }) = (-2,5)$ ; $({ x }_{ 2 },{ y }_{ 2 }) = (2,-3)$ and $({ x }_{ 3 },{ y }_{ 3 }) = (-2,0)$ in the area formula, we get area of triangle $= \left| \dfrac { (-2)(-3+0)+(2)(0-5)+(-2)(5+3) }{ 2 } \right| = \left| \dfrac { 6 - 10 -16 }{ 2 } \right| = \dfrac {20}{2} = 10$ sq. units

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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