If A(-1, 1) and B(2, 3) are two fixed points, then find the locus of

1.	If A(-1, 1) and B(2, 3) are two fixed points, then find the locus of a point P so that the area of triangle APB = 8 sq.units
Answer» 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.

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

Answer»

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.