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Math Help - Co-ordinate Geometry - Applications of a straight line

  1. #1
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    Co-ordinate Geometry - Applications of a straight line

    Hi, guys.

    Can someone please explain to me how to work this out? It's one question out of my homework. I have a bunch of problems similar to this. If someone can explain how to do this I should be able to manage the rest myself.

    "Find the equation of the line joining each of these pairs of given points."
    (4,1) and (7,3)

    Thank you.
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  2. #2
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    The equation of a line is \displaystyle y = mx + c, where \displaystyle m is the gradient, and \displaystyle c is the \displaystyle y intercept.

    First, find the gradient \displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}.

    Then substitute this value, and one of the points, into the equation \displaystyle y = mx + c to find \displaystyle c.
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  3. #3
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    Sorry, I don't fully understand. Could you help me from here?

    (4,1) and (7,3)




    m = (3-1)/(7-4)
    m = 2/3

    y = 2/3(4) + c
    y = 2.66666667 + c <<<?
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  4. #4
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    You also need to let \displaystyle y=1
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    Ahh, still not getting it. The answer is 2x-3y-5=0 but I really don't have a clue how to get there.

    ...

    1 = 2.67 + c
    So c = -1.67?
    Last edited by ebonez; February 22nd 2011 at 09:47 PM.
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  6. #6
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    Quote Originally Posted by ebonez View Post
    Sorry, I don't fully understand. Could you help me from here?

    (4,1) and (7,3)




    m = (3-1)/(7-4)
    m = 2/3

    y = 2/3(4) + c
    y = 2.66666667 + c <<<?
    From the given point (4, 1) substitute x = 4 and y = 1 into \displaystyle y = \frac{2}{3} x + c. Solve for c as an exact fraction (do not use decimal approximations). Then substitute this value back into \displaystyle y = \frac{2}{3} x + c. Now get rid of the fractins by multiplying both sides by 3.

    Show all your working to the above if you're still stuck getting the books final answer.
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