Question

A (5, 3), B (3, −2) are two fixed points; find the equation to the locus of a point P which moves so that the area of the triangle PAB is 9 units.

Answer 1

Let P(h, k) be a point. Let the given points be A(5, 3) and B(3, $-$2).

$$\begin{gathered} \therefore \mathrm{Area}\mathrm{of}∆ABP=\frac{1}{2}\left|\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\}\right| \\ \Rightarrow 9=\frac{1}{2}\left|\left\{5\left(-2-k\right)+3\left(k-3\right)+h\left(3+2\right)\right\}\right| \\ \Rightarrow \left|5h-2k-19\right|=18 \\ \Rightarrow 5h-2k-19=18\mathrm{or}5h-2k-19=-18 \\ \Rightarrow 5h-2k-37=0\mathrm{or}5h-2k-1=0 \end{gathered}$$

Hence, the locus of (h, k) is $5x-2y-37=0\mathrm{or}5x-2y-1=0$.