Question

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

Answer 1

Distance between two points$=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Given:$PQ=QR$

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

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

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

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

squaring both sides, we get

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

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

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

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

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

$\therefore x=-3,5$