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Math Help - Finding the equation of a straight line

  1. #1
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    Smile Finding the equation of a straight line

    Find the equation of the line (in y = mx + c form) that:

    has gradient 2, passing through the intersection of the lines with equations y= 3x - 5 and y= -2x + 5.

    Show any working out...thanks!
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  2. #2
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    Quote Originally Posted by Joker37 View Post
    Find the equation of the line (in y = mx + c form) that has gradient 2, passing through the intersection of the lines with equations y= 3x - 5 and y= -2x + 5.
    To find where the lines intersect, solve the system of equations that they represent.

    Since they're already both solved for "y=", I'd use substitution. So set it up like so:

    . . . . . 3x\, -\, 5\, =\, -2x\, +\, 5

    . . . . . 3x\, +\, 2x\, =\, 5\, +\, 5

    ...and so forth. This will give you the point (x_1,\, y_1) of intersection.

    You are already given the slope (or gradient): m\, =\, 2. Plug this information into whichever straight-line equation you prefer. Since you have a point and a slope, you might want to start with the point-slope form:

    . . . . . y\, -\, y_1\, =\, m(x\, -\, x_1)

    Multiply the slope through the parentheses, and then solve for "y=". The result is the equation you need.

    If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you!
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  3. #3
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    Wink

    3x - 5 = -2x + 5 y = -2(2) + 5
    5x = 10 = -4 + 5
    x = 2 = 1

    y - 1 = 2(x - 2)
    y - 1 = 2x - 4
    y = 2x - 3

    Thanks! Earlier I tried this method but I made a rearragement error.
    i.e at the start I went:

    3x - 5 = -2x + 5
    5x - 25 = 0

    instead what's in the above. Happens when I'm tired!
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