Finding the locus, seriously quick

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The discussion focuses on finding the locus of point P(x,y) given points A(1,4) and B(-3,2) under two conditions: a) angle APB is a right angle and b) P is equidistant from A and the x-axis. For condition a), the relationship M_{PA}M_{PB}=-1 is utilized, where M represents the gradient of the lines PA and PB, leading to the locus equation. For condition b), the distance formula is applied to establish that P's distance from the x-axis (y=0) is equal to its distance from point A.

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1. Given the points A(1,4) and B(-3,2), find the equation of each locus of the pont P(x,y), a) angle APB is a right angle
b) P is equidistant from A and the x-axu
 
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a) The key point is that it's a right angle, so simply use M_{PA}M_{PB}=-1 where M is the gradient of each line. You will end up with an equation in x and y, which is your locus.

b) Since P is equidistant from the x-axis (y=0) and A(1,4) use the distance formula.
 
thanks for your quick response, but could you please explain a) i seem to be having trouble i understanding the equation you have written.
 
Sure, no problem.

Are you familiar with the fact that if two lines are perpendicular to each other, their gradients multiply to give -1?

Such as y=x and y=-x, their gradients multiplied together give: (1)(-1)=-1 and thus these lines are perpendicular (right angles) to each other.

So M_{PA} is the gradient of the line PA, which is found by using the gradient formula,
\frac{y_2-y_1}{x_2-x_1}

If you're still unsure, just ask :smile:
 
Thanks! That makes it a lot more clearer now. Also thanks for your fast responses. =] Hope you have a great day =]
 
Glad I could help :smile: Have a great day too.
 

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