Question

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

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 ⇒ 1 2 12[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 8 ⇒ 1 2 12[x1(1 – 3) + (-1) (3 – y1) + 2(y1 – 1)] = 8 ⇒ 1 2 12[-2x1 – 3 + y1 + 2y1 – 2] = 8 ⇒ 1 2 12[-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.Read more on Sarthaks.com - https://www.sarthaks.com/979576/and-are-two-fixed-points-then-find-the-locus-of-point-so-that-the-area-of-triangle-apb-sq-units

Answer 2

Answer:

Fixed points are A(-1, 1) and B(2, 3).

Given area (formed by these points) of the triangle

APB = 8 ⇒

1/2 12[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 8 ⇒

1/2 12[x1(1 – 3) + (-1) (3 – y1) + 2(y1 – 1)] = 8 ⇒

1/2 12[-2x1 – 3 + y1 + 2y1 – 2] = 8 ⇒

1/2 12[-2x1 + 3y1 – 5] = 8 ⇒

-2x1 + 3y1 – 5 = 16 ⇒

-2x1 + 3y1 – 21 = 0 ⇒

2x1 – 3y1 + 21 = 0

∴ The locus of the point is 2x-3y+21=. 0