Answer :

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


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