Let $A(-1,1)$ and $B(2,3)$ be two points and $P$ be a variable point above the line $A B$ such that the area…
- 6
- $-\frac{6}{5}$
- 4
- $-\frac{12}{5}$
Solution

$\begin{aligned} & \frac{1}{2}\left|\begin{array}{ccc}\mathrm{h} & \mathrm{k} & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1\end{array}\right|=10 \\ & -2 \mathrm{x}+3 \mathrm{y}=25 \\ & -\frac{6}{5} \mathrm{x}+\frac{9}{5} \mathrm{y}=15 \\ & \mathrm{a}=-\frac{6}{5}, \mathrm{~b}=\frac{9}{5} \\ & 5 \mathrm{a}=-6,2 \mathrm{~b}=\frac{18}{5}\end{aligned}$
Asked in: JEE Main 2024 (05 Apr Shift 2)