Question

If A (−1, 1) and B (2, 3) are two fixed points, find the locus of a point P, so that the area of ∆PAB = 8 sq. units.

Answer 1

Let the coordinates of P be (h, k).
Let the given points be $A\left(-1,1\right)\mathrm{and}B\left(2,3\right)$.

$$\begin{gathered} \therefore \mathrm{Area}\mathrm{of}∆PAB=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ \Rightarrow 8\times 2=\left|-1\left(3-k\right)+2\left(k-1\right)+h\left(1-3\right)\right| \\ \Rightarrow 16=\left|-3+k+2k-2-2h\right| \\ \Rightarrow 16=\left|2h-3k+5\right| \\ \Rightarrow 2h-3k+5=16\mathrm{or}2h-3k+5=-16 \\ \Rightarrow 2h-3k-11=0\mathrm{or}2h-3k+21=0 \end{gathered}$$

Hence, the locus of (h, k) is $2x-3y-11=0\mathrm{or}2x-3y+21=0$ .