If $\mathrm{A}=(5, 3), \mathrm{B}(3, -2)$ and a point $\mathrm{P}$ is such that the area of the triangle $\mathrm{PAB}$ is $9 \mathrm{sq}.$ units then the locus of $\mathrm{P}$ represents 

Given vertices are A=(5,3), B=(3, -2)

Let P=(x,y) be locus of third vertex Area of $∆\mathrm{le} \mathrm{PAB}=9$

$$\begin{gathered} \Rightarrow \frac{1}{2}\left|{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)-1{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\right|=9 \\ \Rightarrow \frac{1}{2}\left|5(-2-\mathrm{y})+3(\mathrm{y}-3)+\mathrm{x}(3+2)\right|=9 \\ \Rightarrow \left|-10-5\mathrm{y}+3\mathrm{y}-9+5\mathrm{x}\right|=18 \\ \Rightarrow \left|5\mathrm{x}-2\mathrm{y}-19\right|=18 \\ \mathrm{Squaring} \mathrm{on} \mathrm{b}.\mathrm{s} \\ {(5\mathrm{x}-2\mathrm{y}-19)}^2=(18{)}^2 \Rightarrow 25{\mathrm{x}}^2+4{\mathrm{y}}^2-20\mathrm{xy}-190\mathrm{x}-76\mathrm{y}+37=0 \\ \mathrm{where} {\mathrm{h}}^2={\left(10\right)}^2=100 \\ \mathrm{ab}=25(4)=100 \end{gathered}$$

It represents pair of parallel lines