Question

The midpoint of the sides of a triangle are (5,0), (5,12) and (0,12).The orthocenter of this triangle is

Answer 1

Let triangle be ABC.  A(x1, y1)     B(x2, y2).     C(x3, y3)

Center of AB = D = [ (x1+x2)/2 , (y1+y2)/2 ] = (5, 0)

so    x1 + x2 = 10       y1 = -y2

Similarly,  E = (5, 12) = [(x2+x3)/2, (y2+y3)/2 ] 

  so     x2 + x3 = 10        y2 + y3 = 24

 so      x1 = x3

Further,   F = (0, 12) = [ (x3+x1)/2 , (y3+y1)/2 ] 

  so     x3 + x1 = 0              y3+y1 = 24

 so         x1 = x3 = 0   x2 = 10....

and       y2 = y1 = 0     and y3 = 24

So the vertices:      A(0, 0)     B(10, 0)       C(0, 24)

That is a right angle triangle with sides = 10, 24, 26.

The orthocenter is the origin A(0,0)  itself as the altitudes from B and C meet at A.

HOPE IT HELPS