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# Is this proof suffice as an answer? watch

1. 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).
2. (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
3. 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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