The locus of P such that the area of $∆\mathrm{PAB}$ is 12 sq. units where A(2, 2) and B(–4, 5) is

let $\mathrm{P}=({\mathrm{x}}_1, {\mathrm{y}}_1)$ be locus of point.

Given points A=(2, 2), B=(-4, 5)

Given that Area of ${∆}^{le} \mathrm{PAB}=12$

$$\begin{gathered} \Rightarrow \frac{1}{2}\left|{\mathrm{x}}_1(2-5)+2(5-{\mathrm{y}}_1)-4({\mathrm{y}}_1-2)\right|=12 \\ \Rightarrow \left|-3{\mathrm{x}}_1+10-2{\mathrm{y}}_1-4{\mathrm{y}}_1+8\right|=24 \\ \Rightarrow \left|-3{\mathrm{x}}_1-6{\mathrm{y}}_1+18\right|=24 \\ \Rightarrow \left|{\mathrm{x}}_1+2{\mathrm{y}}_1-6\right|=8 \end{gathered}$$

Squaring on both sides

$$\begin{gathered} {{\mathrm{x}}_1}^2+4{{\mathrm{y}}_1}^2+36+4{\mathrm{x}}_1{\mathrm{y}}_1-24{\mathrm{y}}_1-12{\mathrm{x}}_1=64 \\ \therefore \mathrm{locus} \mathrm{of} \mathrm{P} \mathrm{is} {\mathrm{x}}^2+4\mathrm{xy}+4{\mathrm{y}}^2-12\mathrm{x}-24\mathrm{y}-28=0 \end{gathered}$$