If the point A(2, – 4) is equidistant from P(3, 8)

Solution:

According to the question,

A (2, – 4) is equidistant from P (3, 8) = 0 (-10, y) is equidistant from A (2, – 4)

i.e. $P A=Q A$

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

$\left[\because\right.$ distance between two points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)^{2}}\right]$

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

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

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

On squaring both the sides, we get

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

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

$\Rightarrow \quad y^{2}+8 y+15=0$

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

$\Rightarrow \quad y(y+5)+3(y+5)=0$

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

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

 

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

$\therefore \quad y=-3,-5$

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

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

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

$=\sqrt{169+121}=\sqrt{290}$

Again, distance between $P(3,8)$ and $(-10, y), P Q=\sqrt{(-13)^{2}+(-5-8)^{2}} \quad$ [putting $y=-5$ ]

$=\sqrt{169+169}=\sqrt{338}$

Hence, the values of y are $-3,-5$ and corresponding values of $P Q$ are $\sqrt{2} 90$ and $\sqrt{3} 38=1342$, respectively.