• August 8th 2006, 11:30 PM
chancey
I didn't know which forum to put it in, but I thought this was the most appropriate.

This is the question:
Given two points A(3,-2) and B(-1,7), find the equation of the locus of P(x,y) if the gradient of PA is twice the gradient of PB.

I've tried it 2 different ways and keep getting the same solution:
xy + 5y + 11x - 17 = 0

However the answer in the book says otherwise...

Working:
$PA' = \frac{y+2}{x-3}$

$PB' = \frac{y-7}{x+1}$

$2 PA' = PB'$

$\frac{2y+4}{x-3} = \frac{y-7}{x+1}$

...

$xy + 5y + 11x -17 = 0$
• August 9th 2006, 02:58 AM
galactus
I think you have your 'twice' on the wrong side.

Try $PA'=2PB'$
• August 9th 2006, 04:01 AM
chancey
Quote:

Originally Posted by galactus
I think you have your 'twice' on the wrong side.

Try $PA'=2PB'$

Ahh, it just clicked. I understand why the 2 is on the wrong side. Got the answer, thanks