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)
\[\therefore\text{ Area of ∆ ABP }= \frac{1}{2}\left| \left\{ x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right\} \right|\]
\[ \Rightarrow 9 = \frac{1}{2}\left| \left\{ 5\left( - 2 - k \right) + 3\left( k - 3 \right) + h\left( 3 + 2 \right) \right\} \right|\]
\[ \Rightarrow \left| 5h - 2k - 19 \right| = 18\]
\[ \Rightarrow 5h - 2k - 19 = 18\text{ or }5h - 2k - 19 = - 18\]
\[ \Rightarrow 5h - 2k - 37 = 0\text{ or }5h - 2k - 1 = 0\]
Hence, the locus of (h, k) is
\[5x - 2y - 37 = 0\text{ or }5x - 2y - 1 = 0\]