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 \[A\left( - 1, 1 \right)\text{ and }B\left( 2, 3 \right)\].
\[\therefore\text{ Area of ∆ PAB }= \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]
\[ \Rightarrow 8 \times 2 = \left| - 1\left( 3 - k \right) + 2\left( k - 1 \right) + h\left( 1 - 3 \right) \right|\]
\[ \Rightarrow 16 = \left| - 3 + k + 2k - 2 - 2h \right|\]
\[ \Rightarrow 16 = \left| 2h - 3k + 5 \right|\]
\[ \Rightarrow 2h - 3k + 5 = 16\text{ or }2h - 3k + 5 = - 16\]
\[ \Rightarrow 2h - 3k - 11 = 0\text{ or }2h - 3k + 21 = 0\]
Hence, the locus of (h, k) is
\[2x - 3y - 11 = 0\text{ or }2x - 3y + 21 = 0\]