A(5, 3) and B(3, –2) are two fixed points. Find the equation of locus of P, so that the area of triangle PAB is 9 sq. units.

Given points are A = (5, 3), B = (3, –2)
Let P(x, y) is any point on the locus
Given condition is area of $∆$PAB  9sq.units
area of
$\begin{array}{l}\mathrm{△}\mathrm{PAB}=\frac{1}{2}\left|{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right|\\ \Rightarrow \frac{1}{2}\left|5\right(-2-\mathrm{y})+3(\mathrm{y}-3)+\mathrm{x}(3+2\left)\right|=9\\ \Rightarrow |-10-5\mathrm{y}+3\mathrm{y}-9+5\mathrm{x}|=18\\ \Rightarrow |5\mathrm{x}-2\mathrm{y}-19|=18\Rightarrow 5\mathrm{x}-2\mathrm{y}-19=\pm 18\\ \Rightarrow 5\mathrm{x}-2\mathrm{y}-19=18\text{ or }5\mathrm{x}-2\mathrm{y}-19=-18\\ \Rightarrow 5\mathrm{x}-2\mathrm{y}-37=0\text{ or }5\mathrm{x}-2\mathrm{y}-1=0\end{array}$
locus of P(x, y) is (5x – 2y – 37) (5x–2y–1) = 0