A(2,3) and B(–3, 4) be two given points. Find the equation of locus of P so that area of $∆$PAB is 8.5.

Given points A = (2, 3), B = (–3, 4)
Let P(x, y) be any point on the locus
$\text{ Given condition is area of }\mathrm{△}\mathrm{PAB}=8.5$
$\Rightarrow \frac{1}{2}\left|{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right|=8.5$
$\begin{array}{l}\Rightarrow \frac{1}{2}\left|\mathrm{x}\right(3-4)+2(4-\mathrm{y})-3(\mathrm{y}-3\left)\right|=8.5\\ \Rightarrow |-\mathrm{x}+8-2\mathrm{y}-3\mathrm{y}+9|=17\\ \Rightarrow |-\mathrm{x}-5\mathrm{y}+17|=17\Rightarrow \mathrm{x}+5\mathrm{y}-17=\pm 17\\ \Rightarrow \mathrm{x}+5\mathrm{y}-17=17\text{ or }\mathrm{x}+5\mathrm{y}-17=-17\\ \Rightarrow \mathrm{x}+5\mathrm{y}-34=0\text{ or }\mathrm{x}+5\mathrm{y}=0\end{array}$
$\therefore \text{ The locus of }\mathrm{P}(\mathrm{x},\mathrm{y})\text{ is }$(x + 5y) (x + 5y – 34) = 0