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 of $\triangle \mathrm{PAB}$ is 10 . If the locus of $\mathrm{P}$ is $\mathrm{a} x+\mathrm{b} y=15$, then $5 \mathrm{a}+2 \mathrm{~b}$ is :

• A

  6
• B

  $-\frac{6}{5}$
• C

  4
• Correct answer
  D

  $-\frac{12}{5}$

Solution

$\begin{aligned} & \frac{1}{2}\left|\begin{array}{ccc}\mathrm{h} & \mathrm{k} & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1\end{array}\right|=10 \\ & -2 \mathrm{x}+3 \mathrm{y}=25 \\ & -\frac{6}{5} \mathrm{x}+\frac{9}{5} \mathrm{y}=15 \\ & \mathrm{a}=-\frac{6}{5}, \mathrm{~b}=\frac{9}{5} \\ & 5 \mathrm{a}=-6,2 \mathrm{~b}=\frac{18}{5}\end{aligned}$

Hence, the correct answer is option: 'D'.