Geo2.31

If $(2,1),(-5,7),(-5,-5)\,$ are mid points of sides of a triangle,find the equations of the sides of a triangle.

Let the three vertices are

$A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)\,$

Let the mid point of AB is (2,1)

Hence

$x_1+x_2=4,y_1+y_2=2\,$

Let the mid point of BC is (-5,7)

Hence

$x_3+x_2=-10,y_3+y_2=14\,$

Let the mid point of CA is (-5,-5)

Hence

$x_3+x_1=-10,y_3+y_1=-10\,$

Solving the three six equations we get

$A(2,-11),B(2,13),C(-12,1)\,$

Now the equations are

$y+11=\frac{13+11}{2-2}(x-2),y-13=\frac{1-13}{-12-2}(x-2),y-1=\frac{-11-1}{2+12}(x+12)\,$

$x-2=0,-14y+182=-12x+24,14y-14=-12x-144\,$

$x=2,12x-14y+158=0,12x+14y+130=0\,$

After simplifying,the final equations are

$x=2,6x-7y+79=0,6x+7y+65=0\,$