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    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.
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    (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
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    Would they be in form of y=mx+c?
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    (Original post by Abdo332)
    Would they be in form of y=mx+c?
    Yes

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    Thank you!
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    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)
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    (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 -5=0.5\cdot2+c_1 \Rightarrow c_1=-6

    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.
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    (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 -5=0.5\cdot2+c_1 \Rightarrow c_1=-6

    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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