FULL EXPLANATION ........OTHERWISE GET LOST
ONLY FOR GENIUS
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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 OPTION A{0,0}