# Thread: Locus Question

1. ## Locus Question

The question is as follows:
"The gradient of the line between a moving point P(x,y) and the point A(5,3) is equal to the gradient of the line PB where B has coordinates (2,-1). Find the equation of the locus of P."

Since I haven't done the Locus topic yet I'm really lost...
This is from linear functions...

2. ## Re: Locus Question

If the gradient of PA is equal to the gradient of PB, then P must lie on the line through A AND B. Otherwise the gradients of PA and PB will be different (you can draw up a diagram to visualise this). So in essense, the locus of P is equation of the line through A and B (as P can lie anywhere on this line such that m(PA) = m(PB))

So find the gradient of AB and use the point gradient-formula to find the eqn of the line

m = (-1-3)/(2-5) = 4/3

therefore, y +1 = 4/3(x-2) [rearrange this and you're done]

3. ## Re: Locus Question

Originally Posted by jathu123
If the gradient of PA is equal to the gradient of PB, then P must lie on the line through A AND B. Otherwise the gradients of PA and PB will be different (you can draw up a diagram to visualise this). So in essense, the locus of P is equation of the line through A and B (as P can lie anywhere on this line such that m(PA) = m(PB))

So find the gradient of AB and use the point gradient-formula to find the eqn of the line

m = (-1-3)/(2-5) = 4/3

therefore, y +1 = 4/3(x-2) [rearrange this and you're done]
Thank you so much mate!

4. ## Re: Locus Question

$Find the gradient between point P and A$
$and the gradient between Point P and B$
$m_(AP) = \frac{y-3}{x-5}$
$m_(BP) = \frac{-1-y}{2-x}$
$then the gradients are same$
$m_(AP) =m_(BP)= \frac{y-3}{x-5} = \frac{-1-y}{2-x}$

$After expanding you should get$
$4x-3y-11=0$

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