Question

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

Also find distance $PQ$.

Answer 1

Step 1: State the given data and formula used

It is given that the point $A(2,-4)$ is equidistant from $P(3,8)$ and $Q(-10,y)$.

i.e., $AP=AQ$

According to the distance formula,

The distance $\left(d\right)$ between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ can be calculated by the following formula,

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

$$\begin{gathered} AP=\sqrt{{(3-2)}^2+{(8+4)}^2} \\ \Rightarrow AP=\sqrt{{\left(1\right)}^2+{\left(12\right)}^2} \\ \Rightarrow AP=\sqrt{1+144} \\ \Rightarrow AP=\sqrt{145} \end{gathered}$$

$$\begin{gathered} AQ=\sqrt{{\left(-10-2\right)}^2+{(y+4)}^2} \\ \Rightarrow AQ=\sqrt{{\left(-12\right)}^2+(y^2+16+8y)} \\ \Rightarrow AQ=\sqrt{144+y^2+16+8y} \\ \Rightarrow AQ=\sqrt{y^2+8y+160} \end{gathered}$$

As given, $AP=AQ$

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

$\Rightarrow$ $y^2+8y+160=145$

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

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

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

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

Thus, $y=-5$ or $y=-3$

Step 2: Calculate the distance $PQ$

The distance $PQ$ can be calculated by using the distance formula as,

For $y=-5$

$$\begin{gathered} PQ=\sqrt{{(-10-3)}^2+{(y-8)}^2} \\ \Rightarrow PQ=\sqrt{{(-10-3)}^2+{(-5-8)}^2} \\ \Rightarrow PQ=\sqrt{{(-13)}^2+{(-13)}^2} \\ \Rightarrow PQ=\sqrt{169+169} \\ \Rightarrow PQ=\sqrt{338} \\ \Rightarrow PQ=13\sqrt{2} \end{gathered}$$

For $y=-3$

$$\begin{gathered} PQ=\sqrt{{(-10-3)}^2+{(-3-8)}^2} \\ \Rightarrow PQ=\sqrt{{(-13)}^2+{(-11)}^2} \\ \Rightarrow PQ=\sqrt{169+121} \\ \Rightarrow PQ=\sqrt{290} \end{gathered}$$

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