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    Equations of Lines and Intersections

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    The diagram below (see attachment) shows a triangle ABC whose vertices are at A (-1, 3), B (6, 5) and C (8, -3). The line BP is perpendicular to the line AC, and M is the midpoint of BC.

    Note that BP is called an altitude of triangle ABC and that AM is called a median of triangle ABC.

    a) Find the gradient of
    i) The line AC
    ii) The altitude BP
    (Show answers as a fraction in its simplest form)

    b) Find the equation of the altitude BP, expressing your answer in the form: y = mx + c

    c) Find the coordinates of M.

    d) Find the equation of the median AM, expressing your answer in the form: y = mx + c

    e) Explain how you would find the coordinates of the point of intersection of the altitude BP and the median AM.

    © BrainMass Inc. brainmass.com October 9, 2019, 4:33 pm ad1c9bdddf
    https://brainmass.com/math/triangles/equations-lines-intersections-35782

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    a) Find the gradient of
    gradient of a line (also called slope ) = (y2-y1) /(x2-x1)

    i) The line AC
    Gradient of a line (also called slope ) = (y2-y1) /(x2-x1)

    A (-1,3) , C (8,-3)

    Therefore gradient ={ (-3)-(3) }/{ (8)-(-1)} = -6/9 = -2/3

    Answer: slope of line AC= -2/3
    ii) The altitude BP

    Altitude BP is perpendicular to AC
    The product of the gradients of perpendicular lines = -1
    Therefore ,
    Slope of line AC x slope of line BP = -1
    Or ( -2/3) x slope of line BP = -1
    Or slope of line BP = -1 / (-2/3) = 3 /2

    Answer: slope of line BP= 3/2

    (Show answers as a fraction in its simplest form)

    b) Find the ...

    Solution Summary

    The expert solves equations of lines and intersections. The altitudes of triangles are provided.

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