SSC CHSL 10 Jan 2017 Shift 1 Paper

Let line $\mathit{l}$ perpendicularly bisects line joining A(5,-3) and B(0,2) at C, thus C is the mid point of AB.
Now, slope of AB = $\frac{{\mathit{y}}_2-{\mathit{y}}_1}{{\mathit{x}}_2-{\mathit{x}}_1}=\frac{\left(2+3\right)}{\left(0-5\right)}$
= $\frac{5}{-5}=-1$
Let slope of line $\mathit{l}=\mathit{m}$
Product of slopes of two perpendicular lines = -1
⇒ $\mathit{m}\times -1=-1$
⇒ $\mathit{m}=1$
Equation of a line passing through point $\left({\mathit{x}}_1,{\mathit{y}}_1\right)$ and having slop $\mathit{m}$ is $\left(\mathit{y}-{\mathit{y}}_1\right)=\mathit{m}\left(\mathit{x}-{\mathit{x}}_1\right)$
∴ Equation of line $\mathit{l}$
⇒ $\left(\mathit{y}+\frac{1}{2}\right)=1\left(\mathit{x}-\frac{5}{2}\right)$
⇒ $\mathit{x}-\mathit{y}=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}$
⇒ $\mathit{x}-\mathit{y}=3$