the midpoints of the sides BC, CA and AB of a ΔABC are D(2, 1), B( - 5, 7) and P( - 5, - 5) respectively. Find the equations of the sides of ΔABC.
Let us consider the coordinates of vertices of triangle A, B, C be (a, b), (c, d) and (e, f). Now using mid - point formula
Now from above equations, we have
c + e = 4, d + f = 2 (i)
a + e = - 10, b + f = 14 (ii)
a + c = - 10, b + d = - 10 (iii)
From subtract (i) from (ii),we get
a - c = - 14, b - d = 12 (iv)
Adding (iii) and (iv)
Putting values of a, b in equation (iii)
Again putting values in (i)
So coordinates of A ( - 12,1), B(2, - 11) and C(2,13).
Using two point form of the equation
Equation of side AB:
14(y - 1) = - 12(x + 12)
14y - 14 + 12x + 144 = 0
12x + 14y + 130 = 0
6x + 7y + 65 = 0
Equation of side BC:
y = - 11(slope is not defined i.e. line is vertical)
Equation of side CA:
14(y - 13) = 12(x - 2)
12x - 24 - 14y + 182 = 0
12x - 14y + 158 = 0
6x - 7y + 79 = 0
So, the required equations of sides for AB: 6x + 7y + 65 = 0
For BC: y = - 11
For CA: 6x - 7y + 79 = 0