Question

If $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$, find the values of $x$.

Also find the distance $QR$ and $PR$.

Answer 1

Distance between $QR$ and $PR$:

Given: $Q(0,1)$ is equidistant from $P(5,–3)$and $R(x,6)$ which means $PQ=QR$

Let us find the distance between $PQ$ and $QR$ using distance formula,$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$PQ=\sqrt{{\left(5-0\right)}^2+{\left(-3-1\right)}^2}=\sqrt{25+16}=\sqrt{41}$

$QR=\sqrt{{\left(0-x\right)}^2+{\left(1-6\right)}^2}=\sqrt{x^2+25}$

$$\begin{gathered} PQ=QR \\ \Rightarrow \sqrt{41}=\sqrt{x^2+25} \end{gathered}$$

Squaring both the sides, to omit square root

$$\begin{gathered} 41=x^2+25 \\ {x}^2=16 \\ x=\pm 4 \\ x=4orx=-4 \end{gathered}$$

Coordinates of Point R will be $R(4,6)$OR $R(-4,6)$

If $R(4,6)$, then

$$\begin{gathered} QR=\sqrt{{\left(0-4\right)}^2+{\left(1-6\right)}^2}=\sqrt{41} \\ PR=\sqrt{{\left(5-4\right)}^2+{\left(-3-6\right)}^2}=\sqrt{82} \end{gathered}$$

If $R(-4,6)$ then

$$\begin{gathered} QR=\sqrt{{\left(0-4\right)}^2+{\left(1-6\right)}^2}=\sqrt{41} \\ PR=\sqrt{{\left(5+4\right)}^2+{\left(-3-6\right)}^2}=9\sqrt{2} \end{gathered}$$

Hence the distance $QR=\sqrt{41}$ and $PR=9\sqrt{2}$.