Question

If the point $A(2,-4)$ is equidistant from $P(3,8)$ and $Q(-10,y)$ then find the value of $y$. Also, find distance $PQ$.

Answer 1

Step 1: Apply the distance formula to solve for the value of $y$

According to the question point $A$ is equidistant from points $P$ and $Q$.

$\Rightarrow$ $AP=AQ$

$\Rightarrow$$A{P}^2=A{Q}^2$

Given points: $A(2,-4)$, $P(3,8)$, and $Q(-10,y)$.

The formula to find the distance between two points $\left(x_1,y_1\right)$and $\left(x_2,y_2\right)$is given as

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

Using the distance formula we can write the above equation as

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

$\Rightarrow$ $1+144=144+y^2+8y+16$

$\Rightarrow$ $y^2+8y+15=0$

$\Rightarrow$ $\left(y+3\right)\left(y+5\right)=0$

$\Rightarrow$ $y=-3$ or $y=-5$

Step 2: Solve the distance $PQ$ when $y=-3$

When $y=-3$ , $Q=\left(-10,-3\right)$

Applying distance formula to find $PQ$ we get

$PQ$$=\sqrt{{\left\{3-(-10)\right\}}^2+{\left\{8-(-3)\right\}}^2}$

$=\sqrt{169+121}$

$\Rightarrow$$PQ$ $=\sqrt{290}$ $...\left(i\right)$

Step 3: Solve for the distance $PQ$ when $y=-5$

When $y=-5$, $Q=\left(-10,-5\right)$

Applying the distance formula to find $PQ$ we get

$PQ$$=\sqrt{{\left\{3-(-10)\right\}}^2+{\left\{8-(-5)\right\}}^2}$

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

$\Rightarrow$$PQ$ $=13\sqrt{2}$ $...\left(ii\right)$

Hence, the value of $y$ is $-3$ or $-5$ and the distance $PQ$ is $\sqrt{290}$ units or $13\sqrt{2}$ units respectively for each value of $y$.