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Trig

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How do you find an equation of a line perpendicular to y=x at the point (5,5) in standard form?

  • Trig - ,

    put the original equation in the form
    y = m x + b
    in this case
    y = 1 x + 0
    so m, the original slope, = 1
    Now m' = -1/m
    so our new slope = -1/1 = -1
    so our new function looks like
    y = -1 x + b
    we need b
    use the point
    5 = -5 + b
    b = 10
    so
    y = -x + 10

  • Trig - ,

    The slope of y = x is 1, so the slope of a perpendicular line must be -1. (This is becasue the product of the slopes must be -1).

    The equation can therefore be written
    (y-5)/(x-5) = -1
    y-5 = -x +5
    y = -x +10

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