# Finding the locus, seriously need help quick

1. Apr 2, 2009

### help4drdu

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

2. Apr 2, 2009

### Mentallic

ok for:

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.

3. Apr 2, 2009

### help4drdu

thanks for your quick response, but could you please explain a) i seem to be having trouble i understanding the equation you have written.

4. Apr 2, 2009

### Mentallic

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

5. Apr 2, 2009

### help4drdu

Thanks! That makes it a lot more clearer now. Also thanks for your fast responses. =] Hope you have a great day =]

6. Apr 2, 2009

### Mentallic

Glad I could help Have a great day too.