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Math Help - Perpendicular lines

  1. #1
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    Perpendicular lines

    I can do parallel lines, but not perpendicular

    find the equation of the straight line in form ax + by + c = o which:

    passes through (1,4) and is perpendicular to 5y = x + 2

    thanks
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  2. #2
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    Quote Originally Posted by Carl Feltham View Post
    I can do parallel lines, but not perpendicular
    find the equation of the straight line in form ax + by + c = o which:
    passes through (1,4) and is perpendicular to 5y = x + 2
    If two lines are perpendicular (neither of them is vertical) then the product of their slopes is -1.
    The slope of the given line is \frac {1}{5} so the perpendicular line must have slope -5.
    Carry on.
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  3. #3
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    thanks
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  4. #4
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    Quote Originally Posted by Carl Feltham View Post
    I can do parallel lines, but not perpendicular

    find the equation of the straight line in form ax + by + c = o which:

    passes through (1,4) and is perpendicular to 5y = x + 2

    thanks
    First we need the gradient f the line:

    y = mx + c

    y = 1/5x + 2/5

    Therefore we now know the gradient of the given line is 1/5.

    So the gradient of the perpendicular line must be -5.
    Putting this into the equation of the line we are finding, we get:

    y = -5x + c

    c is the y intercept, which is what we will need to find.

    c = y + 5x

    Now, we just need to put the coordinates into the equation to find c.

    c = 4 + 5(1)

    c = 4 + 5

    c = 9

    We now have our equation

    y = -5x + 9

    (can also be written as: y = 9 - 5x)

    //j0k3r
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