Question

Find the value of y for which the distance between the points P(2, –3) and Q(10, y) is 10 units. (Class 10th)​

Answer 1

• Value of y = –9 and 3.

Given :–

• Two points, P(2, –3) and Q(10, y).
• PQ = 10 units.

To Find :–

• Value of y.

Solution :–

We know that,

$\tt{\blue {\boxed{\red{Distance = \sqrt{{( x_{2} - x_{1})}^{2} {( y_{2} -y_{1})} ^{2} } }}}}$

According to the question,

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

Here,

• $x_{1} = 2$
• $x_{2} = 10$
• $y_{1} = -3$
• $y_{2} = y$
• PQ = 10.

Now, substitute all the values in the distance formula.

$\longmapsto\sqrt{ {(10 - 2)}^{2} {(y - ( - 3))}^{2} } = 10$

$\longmapsto\sqrt{ {(8)}^{2} + {(y + 3)}^{2} } = 10$

Squaring both sides,

$\longmapsto({{\sqrt{ {(8)}^{2} + {(y + 3)}^{2} }}})^{2} = {(10)}^{2}$

Square root cuts the power.

Now, we have

$\longmapsto{(8)}^{2} + {(y + 3)}^{2} = 100$

Apply identity,

$\longmapsto {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}$

So,

$\longmapsto 64 + {(y)}^{2} + 2(y)(3) + {(3)}^{2} = 100$

$\longmapsto 64 + {y}^{2} + 6y + 9 = 100$

$\longmapsto{y}^{2} + 6y + 73 = 100$

$\longmapsto{y}^{2} + 6y + 73 - 100 = 0$

$\longmapsto {y}^{2} + 6y - 27 = 0$

Splitting the middle term.

$\longmapsto{y}^{2} + 9y - 3y - 27 = 0$

$\longmapsto y(y + 9) - 3(y + 9) = 0$

$\longmapsto(y + 9)(y - 3) = 0$

$\longmapsto y + 9 = 0$ or $y - 3 = 0$

$\longmapsto y = -9$ or $y = 3$

Hence,

The value of y is –9 and 3.

Answer 2

Given ,

The distance between the points P(2, –3) and Q(10, y) is 10 units

We know that , the distance between two points is given by

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

Thus ,

$\tt \hookrightarrow 10 = \sqrt{ {(10 - 2)}^{2} + {(y + 3)}^{2} }$

Squaring on both sides , we get

$\tt \hookrightarrow100 = 64 + {(y)}^{2} + 9 + 6y$

$\tt \hookrightarrow {(y)}^{2} + 6y - 27 = 0$

$\tt \hookrightarrow{(y)}^{2} - 9y + 3y - 27 = 0$

$\tt \hookrightarrow y(y - 9) + 3(y - 9) = 0$

$\tt \hookrightarrow(y + 3)(y - 9) = 0$

$\tt \hookrightarrow y = - 3 \: \: or \: \: y = 9$

Hence ,

• Value of of y is either -3 or 9