Equation solving

Perpendicular. To x-3y=2 and contains ( 2,4)


2). That is parallel to 2x+5y=4

I don't get how to solve and use y=Mx+b

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  1. 1.
    given: x - 3y = 2
    -3y = -x + 2
    y = (1/3)x - 2/3
    so the slope of the given line is 1/3
    then the slope of a perpendicular line is -3
    new equation:
    y = 3x + b
    but (2,4) must lie on it, so
    4 = 3(2) + b
    4 = 6 + b
    -2 = b

    new equation: y = 3x - 2

    2. I assume we are dealing with the same point (2,4)
    Easiest way:
    If the new line is parallel to 2x + 5y = 4
    it must differ only in the constant.
    New line: 2x + 5y = c
    with (2,4) on it ...
    2(2) + 5(4) = c = 24

    new equation: 2x + 5y = 24
    To change any equation to the form y = mx + b takes 2 steps:
    1. take all terms except the y term to the right side
    2x + 5y = 24 ------> 5y = -2x + 24
    2. divide each term by the coefficient of the y term
    5y = -2x + 24 ----> y = (-2/5)x - 24/5

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    posted by Reiny

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