Question

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

Answer 1

Answer:

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

Step-by-step explanation:

Given

$A = (-1,-2)$

$B = (2,4)$

$AP:BP = 1 : 2$

Required

The locus of P

$AP:BP = 1 : 2$

Express as fraction

$\frac{AP}{BP} = \frac{1}{2}$

Cross multiply

$2AP = BP$

Calculate AP and BP using the following distance formula:

$d = \sqrt{(x - x_1)^2 + (y - y_1)^2}$

So, we have:

$2 * \sqrt{(x - -1)^2 + (y - -2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}$

$2 * \sqrt{(x +1)^2 + (y +2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}$

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

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