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Thread: need help with stumper

  1. #1
    Sep 2005

    need help with stumper

    How would I write an equation of the line that passes through (-2, 3) and is perpendicular to the line y= 2x +1
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  2. #2
    Senior Member
    Jun 2005
    Let's write lines in the form y = mx+c. If your line passes through the point with x=-2, y=3 then 3 = -2m + c. Now if lines y = mx + c and y = nx + d are perpdendicular then the slopes m and n are related by mn = -1. So your line has slope m = -1/2, being perpendicular to a line with slope 2. That gives you c=2.
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  3. #3
    MHF Contributor
    Apr 2005
    Here is one way.

    You have one point on the unknown line. So if you can find the slope of this line, you can then find this unknown line by using the point-slope form of the equation of a line.
    (y -y1) = m(x -x1) ------***
    where (x1,y1) is the pnown point.

    The unknown line is perpendicular to the known line.
    You know that when two lines are perpendicular, their slopes are negative reciprocals.
    m2 = -1/m1, or m1 = -1/m2 -------***

    In the slope-intercept form of the equation of a line, y = mx +b, the slope of the line is the "m".
    So in the given line,
    y = 2x +1
    its slope is the "2"
    So m1 = 2
    Therefore the slope of the unknown line is m2 = -1/m1 = -1/2.

    Then , the unknown line is
    (y -3) = (-1/2)(x -(-2))
    y -3 = -(1/2)(x +2)
    y -3 = -(1/2)x -1
    y = -(1/2)x -1 +3
    y = -(1/2)x +2 -----answer, in y=mx+b form.

    If you want it in general form Ax +Bx +C = 0,
    y = -(1/2)x +2
    Clear the fraction, multiple both sides by 2,
    2y = -x +4
    x +2y -4 = 0 ------answer also.
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