Question

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

Answer 1

$Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$.
So $PQ=QR$
By distance formula, $\sqrt{\left[\right(5-0{)}^2+(-3-1){)}^2]}=\sqrt{\left[\right(x-0{)}^2+(6-1{)}^2}$
$\sqrt{\left[\right(5{)}^2+(-4^2]}=\sqrt{\left[\right(x^2+\left(5{)}^2\right]}$
$\sqrt{\left[\right(25+16\left)\right]}=\sqrt{[x^2+25]}$
$\sqrt{\left(41\right)}=\sqrt{[x^2+25]}$
Squaring both sides,
$41=x^2+25$
$x^2+25-41=0$
$x^2-16=0$
$(x-4)(x+4)=0$
$(x-4)=0$ or $(x+4)=0$
$x=4$ or $x=-4$
Hence the value of $x$ is $4$ or $-4$.