if orthocentre and circumcentre of a triangleABC is (1,2) , (2,4) respectively.If the equation of side BC is 2x-y-3=0 Then find circum radius of triangle ABC?

I hope the query is resolved.

Solution:- Let coordinates of verteices A, B and C be (x1, y1),(x2, y2) and(x3, y3).

Centroid (G) divides orthocentre (O) and circumcentre (C') in the ratio- 2:1 [Very standard, you should remember this]. So, coordinates of centroid (G) is [(2*0 +9)/3 ,(2*0 +5)/3] = (3, 5/3).

Now, let the perpendicular bisector of line BC be D having coordinates (x,y). Then, y = 2x-10 (since, D lies on line BC). So, D = (x, 2x-10). Also, C'D and BC are perpendicular, which implies-

[(2x-10)/x]*[2] = -1 { i.e., (slope of C'D)*(slope of BC) = -1}

x =5/2 . So,D = (5/2, -5).

(x2+ x3)/2 = 5/2 and(y2+ y3)/2 = -5

or, (x2+ x3) = 5 and(y2+ y3) = -10..........(1)

And, G = (3, 5/3) = [(x1+ x2 +x3)/3,(y1+ y2 +y3)/3]

(x1+ x2 +x3) = 9 and(y1+ y2 +y3) = 5...........(2)

From eq. (1) and (2), we get-

x1 = 4 and y1 = 15

ANS: (4,15)