If the point $ \mathrm{A}(2,-4) $ is equidistant from $ \mathrm{P}(3,8) $ and $ \mathrm{Q}(-10, y) $, find the values of $ y $. Also find distance $ \mathrm{PQ} $.

Given: 

The point \( \mathrm{A}(2,-4) \) is equidistant from \( \mathrm{P}(3,8) \) and \( \mathrm{Q}(-10, y) \).

To do: 

We have to find the values of \( y \) and the distance \( \mathrm{PQ} \).

Solution:

$A(2,-4)$ is equidistant from $P(3,8)$ and $Q(-10, y)$

This implies,

$AP=AQ$

We know that,

The distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is,

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

Therefore,

$\sqrt{(2-3)^{2}+(-4-8)^{2}}=\sqrt{(2+10)^{2}+(-4-y)^{2}}$

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

$\sqrt{1+144}=\sqrt{144+16+y^{2}+8 y}$

$\sqrt{145}=\sqrt{160+y^{2}+8 y}$

Squaring on both sides, we get,

$145=160+y^{2}+8 y$

$y^{2}+8 y+160-145=0$

$y^{2}+8 y+15=0$

$y^{2}+5 y+3 y+15=0$

$y(y+5)+3(y+5)=0$

$(y+5)(y+3)=0$

If $y+5=0$, then $y=-5$

If $y+3=0$, then $y=-3$

The distance between $P(3,8)$ and $Q(-10, y)$ is,

When $y=-3$

$P Q =\sqrt{(-10-3)^{2}+(y-8)^{2}}$

$=\sqrt{(-13)^{2}+(-3-8)^{2}}$

$=\sqrt{169+121}$

$=\sqrt{290}$

When $y=-5$,

$P Q=\sqrt{(-13)^{2}+(-5-8)^{2}}$

$=\sqrt{169+169}$

$=\sqrt{338}$

The values of \( y \) are $-3$ and $-5$. The distance \( \mathrm{PQ} \) is $\sqrt{290}$ when $y=-3$ and $\sqrt{338}$ when $y=-5$.

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Updated on: 10-Oct-2022

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