SSC CHSL 9 Jan 2017 Shift 3 Paper

Coordinates of A (5,0) , B (-2,3) and C (-1,4). Let coordinates of D = (x,y)
ABCD is a parallelogram and AB is parallel to CD, thus slopes of AB and CD are equal.
⇒ $\frac{3-0}{-2-5}=\frac{4-\mathit{y}}{-1-\mathit{x}}$
⇒ $-\frac{3}{7}=\frac{4-\mathit{y}}{-1-\mathit{x}}$
⇒ $3+3\mathit{x}=28-7\mathit{y}$
⇒ $3\mathit{x}+7\mathit{y}=25$ ------(i)
Also, BC is parallel to AD
⇒ $\frac{3-4}{-2+1}=\frac{\mathit{y}-0}{\mathit{x}-5}$
⇒ $\mathit{x}-5=\mathit{y}$ -------(ii)
Substituting value of 'y' in equation (i), we get : $3\mathit{x}+7\left(\mathit{x}-5\right)=25$
⇒ $3\mathit{x}+7\mathit{x}-35=25$
⇒ $\mathit{x}=\frac{60}{10}=6$
Putting it in equation (ii), ⇒ $\mathit{y}=6-5=1$
Now, equation of line AD with coordinates (5,0) and (6,1)
⇒ $\left(\mathit{y}-0\right)=\frac{1-0}{6-5}\left(\mathit{x}-5\right)$
⇒ $\mathit{y}=\mathit{x}-5$