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    Now for b), I'm not certain as to whether this solution is suffice. Is there another way of doing it? If so, please explain.

    Question: Triangle ABC vertices (-2,5) , (6,1) and (2,-7) respectively.

    a) Find the equation of the perpendicular bisector of BC.

    b) Show that this perpendicular bisector passes through the mid-point of side AC.



    Solution:

    a) Midpoint BC = (6+2/2) , (1-7/2) = (4,-3)

    mBC = (-7 - 1) / (2 - 6) = -8/-4 = 2.

    Thus mI = -1/2

    Equation:

    y - b = m (x-a)
    y - (-3) = -1/2 (x-4)
    2(y+3) = -1 (x-4)
    2y + 6 = -1x +4
    2y = -x -2

    Perpendicular Bisector Equation: y = (-1/2 x) - 1


    b) Midpoint AC = (-2+2/2) , (5-7/2) = (0, -1)

    If the perpendicular bisector passes through the midpoint of AC, when the value of x is 0, y = -1 and when the value of y = -1, the value of x = 0, y = -1.

    y = (-1/2)x - 1
    y = (-1/2)(0) - 1
    y = -1

    y = (-1/2)x - 1
    -1 = (-1/2)x - 1
    0 = (-1/2)x
    x = 0.

    As the equation of the perpendicular bisector satisfies the midpoint of AC, the lines must cut through the point (0, -1).
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    (Original post by nerd434)
    Now for b), I'm not certain as to whether this solution is suffice. Is there another way of doing it? If so, please explain.

    Question: Triangle ABC vertices (-2,5) , (6,1) and (2,-7) respectively.

    a) Find the equation of the perpendicular bisector of BC.

    b) Show that this perpendicular bisector passes through the mid-point of side AC.



    Solution:

    a) Midpoint BC = (6+2/2) , (1-7/2) = (4,-3)

    mBC = (-7 - 1) / (2 - 6) = -8/-4 = 2.

    Thus mI = -1/2

    Equation:

    y - b = m (x-a)
    y - (-3) = -1/2 (x-4)
    2(y+3) = -1 (x-4)
    2y + 6 = -1x +4
    2y = -x -2

    Perpendicular Bisector Equation: y = (-1/2 x) - 1


    b) Midpoint AC = (-2+2/2) , (5-7/2) = (0, -1)

    If the perpendicular bisector passes through the midpoint of AC, when the value of x is 0, y = -1 and when the value of y = -1, the value of x = 0, y = -1.

    y = (-1/2)x - 1
    y = (-1/2)(0) - 1
    y = -1

    y = (-1/2)x - 1
    -1 = (-1/2)x - 1
    0 = (-1/2)x
    x = 0.

    As the equation of the perpendicular bisector satisfies the midpoint of AC, the lines must cut through the point (0, -1).
    looks good to me :yy:
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    you should stop stating the obvious. For example, when you plugged in x=0 to show y = -1, you don't have to plug in y = -1 to show x = 0; it is immediately implied first time round. And you should try to be a little less vocal, it saves a lot of time in the exam.

    Otherwise, perfect proof
 
 
 
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