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

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 } } $$
Distance between $$ P (x,y) $$ and $$ A (3,6) = PA = \sqrt { \left( 3-x \right) ^{ 2

}+\left( 6 - y \right) ^{ 2 } } = \sqrt { { 3}^{ 2 }+ { x }^{ 2 } - 6x + { 6}^{ 2 }+ { y }^{ 2 } - 12y }

= \sqrt { { x }^{ 2 } - 6x + { y }^{ 2 } - 12y + 45 } $$

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

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

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

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

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

Distance between $$ (0,y) $$ and $$ (3,1) = \sqrt { \left( 3-0 \right) ^{ 2 }+\left( 1

- y \right) ^{ 2 } } = \sqrt { 9 + { 1}^{ 2 }+ { y }^{ 2 } - 2y } = \sqrt { { y

}^{ 2 } - 2y + 10 } $$

Distance between $$ (0,y) $$ and $$ (1,5) = \sqrt { \left( 1-0 \right) ^{ 2

}+\left( 5 - y \right) ^{ 2 } } = \sqrt { 1 + { 5}^{ 2 }+ { y }^{ 2 } - 10y } =

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

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 $$ (0,2) $$

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 } } $$
Given, Distance between the points $$ (2,3) ; (k,1) = $$ Distance between $$ (2,3) ; (7,k) $$

$$ \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 $$ k = 13 $$

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 $$

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 $$ (0,6) $$

Let the points be $$ A(2,-2), B(-2,1), C(5,2) $$.

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

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

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