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Math Help - Find the distance of the point

  1. #1
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    Find the distance of the point

    Find the distance of the point (-6,-6) from the line through (9,5) which points in the direction of -1I-2J.
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  2. #2
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    The distance the point (p,q) is from the line Ax+By+C=0 equals \frac{{\left| {Ap + Bq + C} \right|}}{{\sqrt {A^2  + B^2 } }}.
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  3. #3
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    Quote Originally Posted by Plato View Post
    The distance the point (p,q) is from the line Ax+By+C=0 equals \frac{{\left| {Ap + Bq + C} \right|}}{{\sqrt {A^2  + B^2 } }}.
    so (p,q) is (-6,-6), how do I figure out Ax+By+C=0?
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  4. #4
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    Quote Originally Posted by viet View Post
    so (p,q) is (-6,-6), how do I figure out Ax+By+C=0?
    That is a problem you need to solve.
    How does one write the equation of a line given a point and a direction vector?
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  5. #5
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    using the parametric equations it should be: x = 9-1t, y= 5-2t
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  6. #6
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    Quote Originally Posted by viet View Post
    using the parametric equations it should be: x = 9-1t, y= 5-2t
    Now write it in the Ax+By+C =0 form.
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