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(-5,3) (5,-3) (6,6) ARE THE MID PIONTS OF A TRIANGLE THEN FIND THE EQUATIONS OF THE SIDES?

Let the vertices of the triangle be A, B and C.

The coordinates of the vertices are: A(ax, ay), B(bx, by), C(cx, cy).

Mid-point of AB is: ((ax + bx)/2, (ay + by)/2) = (-5, 3), say

Mid-point of AC is: ((ax + cx)/2, (ay + cy)/2) = (5, -3), say

Mid-point of BC is: ((bx + cx)/2. (by + cy)/2) = (6, 6), say

Comparing these expressions with the point coordinates given,

ax + bx = -10

ay + by = 6

ax + cx = 10

ay + cy = -6

bx + cx = 12

by + cy = 12

substituting for ax = -bx – 10 and ay = -by + 6 into the bottom 4 eqns,

-bx – 10 + cx = 10  -> -bx + cx = 20

-by + 6 + cy = -6     -> -by + cy = -12

bx + cx = 12

by + cy = 12

Adding together the 1st and 3rd eqns just above and the 2nd and 4th eqns just above gives us,

2cx = 32

2cy = 0

i.e. cx = 16, cy = 0. So the point C is C(16,0)

Substituting for cx = 16 and cy = 0 gives

-bx + cx = 20 -> -bx + 16 = 20 -> bx = -4

-by + cy = -12 -> -by + 0 = -12 -> by = 12

i.e. bx = -4, by = 12. So the point B is B(-4,12)

 

Substituting for cx = 16, cy = 0, bx = -4 and by = 12 gives

ax = -bx – 10 -> ax = 4 – 10 -> ax = -6

ay = -by + 6 -> ay = -12 + 6 -> ay = -6

i.e. ax = -6, ay = -6. So the point A is A(-6, -6)

The vertices of the triangle are: A(-6, -6), B(-4, 12), C(16, 0)

 

The equations of the sides

Side AB

A = (-6, -6), B = (-4, 12)

(y – 12)/(x – (-4)) = (-6 – 12)/(-6 – (-4))

(y – 12)/(x + 4) = -18/(-2) = 9

y – 12 = 9x + 36

y = 9x + 48

Side AC

A = (-6, -6), C = (16, 0)

(y – 0)/(x – 16) = (-6 – 0)/(-6 – 16)

y /(x – 16) = -6/(-22) = 3/11

11y = 3x – 48

Side BC

B = (-4, 12), C = (16, 0)

(y – 0)/(x – 16) = (12 – 0)/(-4 – 16)

y/(x – 16) = 12/(-20) = -3/5

5y = -3x + 48

 

by Level 11 User (81.5k points)

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