NIELIT 2019 Feb Scientist C - Section D: 29

Question

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

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

Answer 1

Answer is C

Answer 2

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