Let the point P(x1, y1).
Fixed points are A(-1, 1) and B(2, 3).
Given area (formed by these points) of the triangle
APB = 8
⇒ \(\frac{1}{2}\)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 8
⇒ \(\frac{1}{2}\)[x1(1 – 3) + (-1) (3 – y1) + 2(y1 – 1)] = 8
⇒ \(\frac{1}{2}\)[-2x1 – 3 + y1 + 2y1 – 2] = 8
⇒ \(\frac{1}{2}\)[-2x1 + 3y1 – 5] = 8
⇒ -2x1 + 3y1 – 5 = 16
⇒ -2x1 + 3y1 – 21 = 0
⇒ 2x1 – 3y1 + 21 = 0
∴ The locus of the point P(x1, y1) is 2x – 3y + 21 = 0.