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The scalar product

  1. Sep 6, 2006 #1
    Today I was given a problem to solve in class and was told to complete it for homework. This problem is as follows:

    The line y=mx + c has a gradient m and cuts the y axis at (0,c). Thus we can write the parametric vector equation of the line as:

    [tex]r = cj +\lambda (i + mj)[/tex]

    Using this fact show that that the perpendicular distance from point [tex]A(x_1 , y_1)[/tex] to y = mx + c is:

    [tex]\mid(\frac{mx_{1} - y_{1} + c}{\sqrt{m^2 + 1}})\mid[/tex]

    If y = mx + c is instead written as ax + by + d = 0 show that the perpendicular distance of point [tex]A(x_1 , y_1)[/tex] to as ax + by + d = 0 is given by:

    [tex]\mid(\frac{ax_{1} - by_{1} + d}{\sqrt{a^2 + b^2}})\mid[/tex]

    This diagram which I drew to help me may help:



    I have tried solving this problem by using vectors:


    and I know that the dot product of [tex]( x_1 , y_1 )[/tex] and y = mx + c is equal to zero but from there onwards I am not sure on how to approach this problem. All help is appreciated,
    thanks, Pavadrin
  2. jcsd
  3. Sep 6, 2006 #2
    First, you need to find a vector perpendicular to the line y = mx + c. Call this vector u. Let v be the vector that points to A. Then for some n, nu + v = r. The magnitude of nu is the perpendicular distance from the point A to the line y = mx + c.
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