Question

The mid point of the sides of the triangle are (5,0),(5,12) and (0,12).the orthocenter of the 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.

Answer 2

ANSWER

 

  Let P (x1, y1), Q(x2, y2) and R(x3, y3) be the coordinates of the vertices of the given triangle, then

let center of PQ = D

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

Hence we get

=> X1+X2 =10 and Y1=(-Y2)              --------------(i)

and let the center of QR = E

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

Hence we have

=>x2 + x3 = 10   and  y2 + y3 = 24                 ---------(ii)

So from (i) and (ii) we get-->  x1 = x3                 -----------------(iii)

let the center of  be RP F

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

hence we get=>

   x3 + x1 = 0    and    y3+y1 = 24                    ------------------(iv)

From (i),(ii) ,(iii) and (iv) we get

 -    x1 = x3 = 0   and  x2 = 10

 -    y2 = y1 = 0   and  y3 = 24

so the vertices of the triangle ABC are =>

A(0, 0), B(10, 0)  and C(0, 24

But this is a  right angle triangle with sides which is having sides as  10units, 24units and 26units respectively

hence we can see that orthocenter of the points is the origin A(0,0)  itself as the altitudes from B and C meet at A.

So we get that the answer is {0,0}