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Math Help - Finding the equation of the straight line

  1. #1
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    Finding the equation of the straight line

    Diagram shows a kite PQRS.Points P and R lie on Y-axis and X-axis respectively.Point S is

    (-1,5) Q(6,12)



    a)Given PS = PQ, Find

    i)the equation of PR
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  2. #2
    MHF Contributor Unknown008's Avatar
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    First, find the coordinates of the point P.

    Let it be (x, y)

    The distances PS and PQ are given by:

    |PS| = \sqrt{(12-y)^2 + (6-x)^2}

    |PQ| = \sqrt{(5-y)^2 + (-1-x)^2}

    Since they are equal;

    \sqrt{(5-y)^2 + (-1-x)^2} = \sqrt{(12-y)^2 + (6-x)^2}

    (5-y)^2 + (-1-x)^2 = (12-y)^2 + (6-x)^2

    Now, you know that the x coordinate of P is zero... it becomes simpler!

    (5-y)^2 + (-1-0)^2 = (12-y)^2 + (6-0)^2

    (5-y)^2 + 1 = (12-y)^2 + 36

    Finish that to get the y coordinate of P.

    Find the midpoint of QS. I assume you know how to find that.

    The line through MP (where M is the midpoint of QS) also passes through R. You got two points (P and M). You can find the equation of PR.

    OR!

    Find the midpoint of QS. (1)

    Find the gradient of QS. (2)

    Find the normal of the gradient of QS. (3)

    You get a gradient (in (3)) and a point (in (1)). Find the line through PR!

    This method is shorter and makes use the the property in kites that their diagonals are perpendicular.
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  3. #3
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    mid point of QS i get \left(\frac{5}{2},\frac{17}{2}\right) normal i

    gradient QS = 1

    the normal is -1

    therefore,the equation is

    y- \frac{17}{2} = -1\left(x-\frac{5}{2}\right)

    y= -x + 11

    right?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Correct!
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  5. #5
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    so,the point p is (0,11)?

    and the point r is?
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  6. #6
    MHF Contributor Unknown008's Avatar
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    You know that R has its y coordinates as 0. So, put y = 0 and x = ??
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  7. #7
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    owh..

    point r =(11,0)

    point p=(0,11)?
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Right!
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