Question

$A(5,3)$ 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.

Answer 1

Let $P(h,k)$

$Ar△PAB=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=9$

$P\equiv (h,k)\equiv (x_1,y_1)$

$A\equiv (5,3)\equiv (x_2,y_2)$

$B\equiv (3,-2)\equiv (x_3,y_3)$

$\therefore Ar△PAB=\frac{1}{2}\left|\begin{array}{ccc}h& k& 1\\ 5& 3& 1\\ 3& -2& 1\end{array}\right|=9$

upon simplification we get,

$5h-2k=37$

$\Rightarrow 5x-2y=37$ is the required equation.