Question

Given points A(-1, -2) and B(2, 4) where AP: BP=1:2, find the locus of point P.​

Answer 1

`x^2 + 4x + y^2 +8y = 0`

Explanation:

Given

`A = (-1,-2)`

`B = (2,4)`

`AP:BP = 1 : 2`

Required

The locus of P

`AP:BP = 1 : 2`

Express as fraction

`(AP)/(BP) = (1)/(2)`

Cross multiply

`2AP = BP`

Calculate AP and BP using the following distance formula:

`d = √((x - x_1)^2 + (y - y_1)^2)`

So, we have:

`2 * √((x - -1)^2 + (y - -2)^2) = √((x - 2)^2 + (y - 4)^2)`

`2 * √((x +1)^2 + (y +2)^2) = √((x - 2)^2 + (y - 4)^2)`

Take square of both sides

`4 * [(x +1)^2 + (y +2)^2] = (x - 2)^2 + (y - 4)^2`

Evaluate all squares

`4 * [x^2 + 2x + 1 + y^2 +4y + 4] = x^2 - 4x + 4 + y^2 - 8y + 16`

Collect and evaluate like terms

`4 * [x^2 + 2x + y^2 +4y + 5] = x^2 - 4x + y^2 - 8y + 20`

Open brackets

`4x^2 + 8x + 4y^2 +16y + 20 = x^2 - 4x + y^2 - 8y + 20`

Collect like terms

`4x^2 - x^2 + 8x + 4x + 4y^2 -y^2 +16y + 8y + 20 - 20 = 0`

`3x^2 + 12x + 3y^2 +24y = 0`

Divide through by 3

`x^2 + 4x + y^2 +8y = 0`