Priyansh Bajaj AskiitiansExpert-IITD
Last Activity: 14 Years ago
Dear Aman,
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)
Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Aman!!!
Regards,
Askiitians Experts
Priyansh Bajaj