(-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