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Math Help - GCSE Question

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    Member GAdams's Avatar
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    GCSE Question

    [IMG]file:///C:/DOCUME%7E1/SHAKEE%7E1/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]


    Can anybody help me with this question please?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    [IMG]file:///C:/DOCUME%7E1/SHAKEE%7E1/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]


    Can anybody help me with this question please?
    First let's find the point M

    The formula for the midpoint is given by:

    \left( \frac {x_1 + x_2}{2}, \frac {y_1 + y_2}{2} \right)

    Using (x_1,y_1) = A(2,9) and (x_2, y_2) = B(8,7), the point M is:

    M = \left( \frac {2 + 8}{2}, \frac {9 + 7}{2} \right) = (5,8)

    Now we can find the slope of the lines connecting M and C, and the slope of the line connecting A and B, if there slopes are the negative inverses of each other (that is, the product of their slopes is -1), then MC is perpendicular to AB.

    Let m_1 be the slope of the line connecting AB
    Let m_2 be the slope of the line connecting MC

    Now, using (x_1,y_1) = (2,9) \mbox { and } (x_2,y_2) = (8,7)

    \Rightarrow m_1 = \frac {y_2 - y_1}{x_2 - x_1} = \frac {7-9}{8-2} = - \frac {1}{3}


    Now, using (x_1,y_1) = (5,8) \mbox { and } (x_2 , y_2) = (8,18)

    \Rightarrow m_2 = \frac {18 - 8}{8 - 5} = \frac {10}{3}

    Thus, MC and AB are not perpendicular
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