# 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 $\mathrm{PQ}$.

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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 $y=-5$.

Updated on 10-Oct-2022 13:28:51