SSC CHSL 7 Jan 2017 Shift 3 Paper

Diagonals of a rhombus bisect each other perpendicularly. Thus, O is the mid point of QS and RP.
Coordinates of midpoint of line joining A = $\left({\mathit{x}}_1,{\mathit{y}}_1\right)$ and B = $\left({\mathit{x}}_2,{\mathit{y}}_2\right)$ is $\left(\frac{{\mathit{x}}_1+{\mathit{x}}_2}{2},\frac{{\mathit{y}}_1+{\mathit{y}}_2}{2}\right)$
⇒ O = $\left(\frac{3-7}{2},\frac{1+5}{2}\right)=\left(-2,3\right)$
Slope of PR = $\frac{1-5}{3+7}=-\frac{2}{5}$
Product of slopes of two perpendicular lines is -1
⇒ Slope of QS = $\mathit{m}\times -\frac{2}{5}=-1$
⇒ $\mathit{m}=\frac{5}{2}$
∴ Equation oof line QS with slope $\mathit{m}=\frac{5}{2}$ and passing through point O $\left({\mathit{x}}_1,{\mathit{y}}_1\right)=\left(-2,3\right)$
$=\left(\mathit{y}-{\mathit{y}}_1\right)=\mathit{m}\left(\mathit{x}-{\mathit{x}}_1\right)$
$=\left(\mathit{y}-3\right)=\frac{5}{2}\left(\mathit{x}+2\right)$
$=2\mathit{y}-6=5\mathit{x}+10$
$=5\mathit{x}-2\mathit{y}=-16$