Question

If $(-4,0)$ and $(1,-1)$ are two vertices of a triangle of area $4$ sq. units, then its third vertex lies on

• A. $y=x$
• B. $5x+y+12=0$
• C. $x+5y-4=0$
• D. $x+5y+12=0$

Answer 1

Answer: (C), (D)

The correct options are
C $x+5y-4=0$
D $x+5y+12=0$

Let the third vertex be $(x,y)$. Then,
$\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ -4& 0& 1\\ 1& -1& 1\end{array}\right|=\pm 4$
$\Rightarrow x+5y+4=\pm 8$
$\Rightarrow x+5y+12=0orx+5y-4=0$
Hence, the third vertex lies on $x+5y+12=0$ or $x+5y-4=0$