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Math Help - coordinate geometry

  1. #1
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    coordinate geometry

    Find the perpendicular distance between the pair of parallel lines 3x-4y+12 = 0 and 3x-4y+2=0


    How do i start ??
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  2. #2
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    Quote Originally Posted by mathaddict View Post
    Find the perpendicular distance between the pair of parallel lines 3x-4y+12 = 0 and 3x-4y+2=0


    How do i start ??
    With your question there are two different ways to find the answer:

    1. Choose any arbitrary point on the 1rst line; determine the equation of a line perpendicular to the 1rst line passing through this point; calculate the intersection point of the perpendicular line and the 2nd line; calculate the distance between the choosen point and the intersection point.

    2. The equations of the two lines are given in normal form:

    3x-4y+12=0~\implies~(3, -4)\cdot (x, y) +12 = 0 with

    (3, -4) indicating the perpendicular direction to the line and

    12 indicating the perpendicular distance of the origin to the line.

    If the normal vector is a unit vector then the distance of origin to the line is measured in units too. The difference between these two distances must be the distance between the two parallels:

    |(3, -4)| = 5 Therefore the equation of the two lines become:

    \frac35x-\frac45y+\frac{12}5=0\ and\ \frac35x-\frac45y+\frac{2}5=0

    Therefore the distance between the parallels is:

    d = \frac{12}5 - \frac25 = 2
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