Question

The $x$-coordinate of a point $P$ is twice its $y$-coordinate. If $P$ is equidistant from $Q(2,-5)$and $R(-3,6)$, find the coordinates of $P$.

Answer 1

Formula for distance of line$=\sqrt{{\left(x-x_1\right)}^2+\left(y-y_1\right)2}$
Let $(x,y)$ be the coordinates of $P$.

Given that $x=2y$

Also,

$PQ=PR$
$\sqrt{{\left(x-2\right)}^2+{\left(y+5\right)}^2}=\sqrt{{\left(x+3\right)}^2+{\left(y-6\right)}^2}$
${\left(x-2\right)}^2+{\left(y+5\right)}^2={\left(x+3\right)}^2+{\left(y-6\right)}^2$ (squaring both sides)
Put $x=2y$ in equation:
$$\begin{gathered} {\left(2y-2\right)}^2+{\left(y+5\right)}^2={\left(2y+3\right)}^2+{\left(y-6\right)}^2 \\ 4y^2+4-4y+y^2+25+10y=4y^2+9+6y+y^2+36-12y \\ 2y+29=45 \\ 2y=16 \\ y=8 \end{gathered}$$
Hence, $x=2y=2\times 8=16$
Therefore, the coordinates of $P$ are $(16,8)$.