Question

the mid point of the sides of a triangle are (2,2),( 2,3) and (4,6) . find the equation of the sides of the triangle​

Answer 1

Answer:

Let ABC be the triangle such that:

mid point AB = ((xa+xb)/2, (yb+yb)/2) = (2,2)

==> xa+xb/2 = 2==> xa+xb = 4......(1)

==> (ya+yb)/2 = 2 ==> ya+yb = 4.......(2)

midpoint AC= (xa+xc)/2, (ya+yc,2) = (2,3)

==> (xa+xc)/2 =2 ==> xa+xc = 4 .......(3)

==> (ya+yc)/2 = 3==> ya+yc = 6........(4)

midpoint BC = ((xb+xc)/2, (yb+yc)/2) = (4,6)

==> (xb+xc)/2 = 4 ==> xb+xc = 8..........(5)

==> (yb+yc)/2 = 6==> yb+yc = 12..........(6)

Now to solve:

Let us solve the system with equation (1), (3) , and (5)

xa+xb = 4 .........(1)

xa +xc = 4..........(2)

xb+xc = 8 .........(3)

Subtract (1) from (2)

==> xc - xb = 0 ==> xc = xb

==> xc + xc = 8

==> xc=4, xb =4, xa= 0

Now let us solve the system with equations (2), (4), and (6)

ya+yb = 4.......(2)

ya + yc = 6 .......(4)

yb + yc = 12.......(6)

Subtract (2) from (4)

==> yc-yb = 2 .....(7)

now add (7) and (6)

==> 2yc = 14

==> yc = 7 , yb = 5, ya = -1

Then A(0, -1) , B(4,5) , and c(4,7)