If $(-4,0),(1,-1)$ are two vertices of a triangle whose area is $4$ Sq units then its third vertex lies on:

Answer is C.

Area = $\frac{1}{2} determinant\begin{vmatrix} -4&0&1 \\ 1&-1&1 \\ x&y&1 \end{vmatrix} = 4$

After expanding, we will get x + 5y – 4 =0