0 votes 0 votes If $(-4,0),(1,-1)$ are two vertices of a triangle whose area is $4$ Sq units then its third vertex lies on: $y=x$ $5x+y+12=0$ $x+5y-4=0$ $x-5y+4=0$ Quantitative Aptitude nielit2019feb-scientistc quantitative-aptitude geometry + – Lakshman Bhaiya asked Apr 1, 2020 • recategorized Nov 8, 2020 by Krithiga2101 Lakshman Bhaiya 13.7k points 597 views answer comment Share See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Answer is C s_dr_13 answered Dec 11, 2020 s_dr_13 228 points comment Share See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 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 neethu_seb answered Dec 10, 2022 neethu_seb 3.4k points comment Share See all 0 reply Please log in or register to add a comment.