Question

A curve is traced by a point P(x, y) which moves such that its distance from the point A(-1,1) is three times its distance from the point B(2,-1). Determine the equation of the curve.

Answer 1

Answer:

$8x^2+8y^2+43-38x+20y=0$

Step-by-step explanation:

Let $A\left ( x_1,y_1 \right )$ and $B\left ( x_2,y_2 \right )$ be two points then distance AB is equal to $AB=\sqrt{\left ( x_2-x_1 \right )^2+\left ( y_2-y_1 \right )^2}$

Here, a curve is traced by a point $P(x,y)$ which moves such that its distance from the point $A(-1,1)$ is three times its distance from the point $B(2,-1)$ i.e $AP=3BP$

Using distance formula,

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

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

$AP=3BP\\\sqrt{(-1-x)^2+(1-y)^2}=3\sqrt{(2-x)^2+(-1-y)^2}$

On squaring both sides, we get

$(-1-x)^2+(1-y)^2=9\left [ (2-x)^2+(-1-y)^2 \right ]\\1+x^2+2x+1+y^2-2y=9\left ( 4 +x^2-4x+1+y^2+2y\right )\\1+x^2+2x+1+y^2-2y=36+9x^2-36x+9+9y^2+18y\\8x^2+8y^2+43-38x+20y=0$

So, equation of curve is $8x^2+8y^2+43-38x+20y=0$