If $P,Q,R$ are $\left(6,-1\right),\,\left(1,3\right),\,\left(x,8\right)$ respectively, then find the value of $x$ so that $PQ=QR$

Distance between two points$=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

Given:$PQ=QR$

$\Rightarrow \sqrt{{\left(1-6\right)}^{2}+{\left(3+1\right)}^{2}}=\sqrt{{\left(x-1\right)}^{2}+{\left(8-3\right)}^{2}}$

$\Rightarrow \sqrt{{\left(-5\right)}^{2}+{\left(4\right)}^{2}}=\sqrt{{\left(x-1\right)}^{2}+{\left(5\right)}^{2}}$

$\Rightarrow \sqrt{25+16}=\sqrt{{x}^{2}-2x+1+25}$

$\Rightarrow \sqrt{41}=\sqrt{{x}^{2}-2x+26}$

squaring both sides, we get

$\Rightarrow {x}^{2}-2x+26-41=0$

$\Rightarrow {x}^{2}-2x-15=0$

$\Rightarrow {x}^{2}+3x-5x-15=0$

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

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

$\therefore x=-3,\,5$