Q. 23

# the midpoints of

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)

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

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