Question

find the equation of the locus of a point which forms a triangle of area 2 with the point A(1,1) B(-2,3)

Answer 1

Let P ≡ (x ,y) is the locus of point which forms a triangle of area 2 sq unit with points A (1,1) and B(-2,3).

We know, area of triangle formed by points (x₁,y₁) ,(x₂, y₂) and (x₃,y₃).
Then, area of triangle = 1/2 [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

So, area of triangle Formed by P(x,y) , A(1,1) and B(-2,3)
2 = 1/2 |x(1 - 3) + 1(3 - y) -2(y - 1)|
⇒ 4 = |-2x + 3 - y -2y + 2 |
⇒ 4 = |-2x -3y + 5|
breaking mod
⇒±4 = (-2x -3y + 5)
⇒ -2x - 3y + 5 = ± 4
⇒-2x - 3y = 5 ± 4

Hence , equations of locus of point are 2x + 3y + 9 = 0
and 2x + 3y + 1 = 0

Answer 2

Answer:

let p(x,y)be the third vertex of∆PAB

Given that A=(1,1)

B=(-2,3)

Area of a triangle PAB

1/2|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|=2

1/2|1(3-y-2(y-1)+(1-3)|=2

|3-1y-2y+2+1x-3x|=2×2

|2x-3y+5|=4

2x-3y+5=+or-4

2x-3y+5=4

2x-3y+5-4=0

2x-3y+1=0

2x-3y+5=-4

2x-3y+5+4=02x-3y+9=0

Required equation of locus is

(2x-3y+1) (2x-3y+9)=0