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# Coordinate Geometry Question? watch

1. hey guys. I have a question I'm stuck on.

The question is: Three points have coordinates A (2,9) B (4,3) C (2,-5). the line through C with gradient .5 meets the line AB produced at D. Calculate the coordinates of D.

I've started with drawing up a graph and plotting the points, however I do not know what to do afterwards to find the point of intersection D.

sorry for a pretty simple question, Maths isn't my strong point.
2. (Original post by Abdo332)
hey guys. I have a question I'm stuck on.

The question is: Three points have coordinates A (2,9) B (4,3) C (2,-5). the line through C with gradient .5 meets the line AB produced at D. Calculate the coordinates of D.

I've started with drawing up a graph and plotting the points, however I do not know what to do afterwards to find the point of intersection D.

sorry for a pretty simple question, Maths isn't my strong point.
Construct a pair of simultaneous equations and solve.
One will be for the line with gradient 0.5 and the other will be for the line including AB
3. Would they be in form of y=mx+c?
4. (Original post by Abdo332)
Would they be in form of y=mx+c?
Yes

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5. Thank you!
6. I'm really not having a frickin clue as how to set this up for whatever reason.

So one of them is y= .5x + c
So I use the points of C for (X,Y)? So it becomes -5=.5(2) + C?

And the other line I found the gradient, so it's now y= -3x+C?
Again, do I use points from A & B or (X,Y)
7. (Original post by Abdo332)
I'm really not having a frickin clue as how to set this up for whatever reason.

So one of them is y= .5x + c
So I use the points of C for (X,Y)? So it becomes -5=.5(2) + C?

And the other line I found the gradient, so it's now y= -3x+C?
Again, do I use points from A & B or (X,Y)
So for the line passing through C, you have a line gradient 0.5, so it is of the form y=0.5x+c1
To work out the c1 you need to substitute the values of x and y for a point you know to be on the line, in this case, (2,-5); so you get

If you then follow a similar procedure to get the line joining A and B, first having to find the gradient which, since it's a straight line, can be found by taking the difference in the y value of the two points and divide it by the difference in the x value. Then do exactly the same as above to find an equation for that line.

At this point you should have two equations of the form y=a1x+c1 and y=a2x+c2. You can then solve these simultaneous equations to find the x value of the intercept and then plug that into one of the two formulae that you found to get the y value.
8. (Original post by Jammy Duel)
So for the line passing through C, you have a line gradient 0.5, so it is of the form y=0.5x+c1
To work out the c1 you need to substitute the values of x and y for a point you know to be on the line, in this case, (2,-5); so you get

If you then follow a similar procedure to get the line joining A and B, first having to find the gradient which, since it's a straight line, can be found by taking the difference in the y value of the two points and divide it by the difference in the x value. Then do exactly the same as above to find an equation for that line.

At this point you should have two equations of the form y=a1x+c1 and y=a2x+c2. You can then solve these simultaneous equations to find the x value of the intercept and then plug that into one of the two formulae that you found to get the y value.

thanks heaps!

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