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Mathematics Help -Straight line graphs

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Tue Aug 28, 2012 8:58 am
laylaflame says...

Hey guys, if anyone could help with these few questions about straight line graphs, it would be amazingly wonderful:D
I'm in yr11 Advanced Maths, and it would great if you could explain how to answer these questions, however I will give a bit of background on each of them.
There is about 4 questions...

1.Midpoints on a straightline:
-using formula:
Midpoint = ( x(1) + x(2) /2, y(1) + y(2) /2)

Q. a) Point P (x,y) moves so that the midpoint between P and the origin is always a point on the circle x^2 + y^2 =1. Find the equation of the locus of P.

(I don't understand finding the equation of a locus)

Q. b) Find the equation of the locus of the point P(x,y) that is the midpoint between all points on the circle x^2 + y^2 =4 and the origin.

2. Gradient (m)
- m= y(2)- y(1) / x(2) - x(1)
- m= rise/ run

(again this involves locus)
Q. 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.

3. This is do to with finding straight line equations-
Q. A straight line has x-intercelt of 4 and passes through (0,-3). Find its equation.

Also.. I can't remember what collinear is... :)

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Mon Sep 03, 2012 10:28 pm
artsy says...

I know this doesn't help much, but collinear is basically like lying on the same straight line.

Sorry, we haven't covered locus in my math class, and even if we had, I would be having the same problem too.

Hope you find what you're looking for!
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Wed Jan 16, 2013 8:25 pm
Cadi says...

Hey there, layla,

I know it's been a long time since you posted this (I hope you got some help sooner than this from somewhere!), but I figured I would post an answer, just in case someone else is having the same problem and wound up here.

Okay, so a locus is a set of points - that means, it can be a line (straight or curved) or an area. So the locus of P is all the places where P could be, given the information you have. In this case, the locus of P is going to be a line, probably a curved one, so when we "find the equation of the locus", we're looking to find an equation of a curve.

There's information here about the midpoint between P and the origin - the point halfway between (0,0) and wherever P is. Let's call the midpoint Q, to save on writing. Now, we know the equation of the locus of Q (the equation of the curve that is places Q can be) - because it's given to us. x^2 + y^2 = 1.

We also know, thanks to the midpoint formula, how to find a midpoint from two endpoints. The x co-ordinate of the midpoint is the sum of the endpoint x co-ords, divided by 2 - and similarly for the y co-ordinate. Remember, Q is our midpoint, and our two endpoints are (0,0) and P. That means we can clear up the midpoint formula a little bit, by replacing x(1) and x(2) with x(P) [which is a placeholder for the x value of P that we don't know yet] and 0. So we end up with the midpoint being at (x(P)/2, y(P)/2).

Now, because the equation for the locus of Q (with the 'squared's in) and the midpoint formula are both talking about the x and y values of Q, we can take the midpoint formula and substitute it into the locus of Q. I'll try to write this out just below, but it looks nasty in typed text - I'm sorry ><

(1/2 * x(P))^2 + (1/2 * y(P))^2 = 1 [* means multiply, I can't remember what year of school that gets introduced]

Now, if we do the squaring, we get...

(1/4 * x(P)^2) + (1/4 * y(P)^2) = 1

...which you can make look nicer by multiplying both sides by 4...

x(P)^2 + y(P)^2 = 4

...and tadaa. This is an equation of a curve (a circle, actually), which represents all the places the point P could be.

This is a similar process to 1a - work out what the endpoints will be, and use the midpoint formula together with the other equation given to work out the locus you need. A lot of the time, it's also useful to draw a diagram - again, something that's tricky for me to demonstrate on the computer, but if you just make a quick sketch of the information you have, it can be easier to work out what's going on.

Again, this is a similar kind of thing - we're trying to find an equation involving the co-ordinates of P, and we're going to get to it by substituting things we know into other things we know. The difference is, the things we know in this question are about gradients of lines (the 'm' in those equations) and gradients being the same as each other.

So first, looking at the line between P and A. Using that gradient equation, we can put in the x and y values for P and A, and get back m(PA) = (y(P) - 3) / (x(P) - 5)

Doing the same thing with P and B, we get m(PB) = (y(P) - 2) / (x(P) + 1).

The other thing the question tells us is that the gradient of the line between P and A is the same as the gradient of the line between P and B - so m(PA) needs to be the same number as m(PB). That means we can take the right hand side of our two gradient equations (for m(PA) and m(PB)) and make them equal to each other:

(y(P) - 3) / (x(P) - 5) = (y(P) - 2) / (x(P) + 1)

Now we have an equation in terms of the co-ordinates of P, which is what we were after. It looks a bit nasty, though, so we can rearrange it a bit to look better:

(y(P) - 3) * (x(P) + 1) = (y(P) - 2) * (x(P) - 5)

y(P)*x(P) - 3*x(P) + y(P) - 3 = y(P)*x(P) - 2*x(P) - 5*y(P) + 10

-3*x(P) + y(P) - 3 = -2*x(P) - 5*y(P) + 10

6*y(P) - x(P) - 7 = 0

Last question! The equation of a line is y = mx + c, where m is the gradient (as we were using in Q2) and c is the place it crosses the y axis. So we can find the c easily for this question - it's given in the (0,-3) co-ordinate. And the gradient is also easily - the equation in Q2 means we can find m if we have two points on the line, and because we're told the x-intercept (the place it crosses the x axis) and the y-intercept (the (0,-3)), we have those two points and can just plug them into the equation!

Very finally, 'co-linear' just means 'on the same line' - so A, B and C are co-linear points if you can draw a straight line through all three.
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