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```				   Let ABC be a triangle having orthocentre & circumcentre at (9,5) and (0,0) respectively. If the equation of side BC is 2x-y=10,then find the possible coordinates of vertex A.
```

6 years ago

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```										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 questionDifficult 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 ExpertsPriyansh Bajaj
```
6 years ago
```										I think there is a mistake in above answer....
Hmm... so as calculated above centroid comes out to be (3,5/3)
now   x1 + x2 +x3 = 9   -(1)    and y1 + y2 + y3 = 5      (2)  ,
now see, line joining the othocenter and (x1,y1) is perpendicular to the base 2x - y = 10, so slope of the line will be -1/2
we know it will pass through (9,5) ie the orthocenter so, using point slope form, we have equation of line as 2y + x = 19 - (3)
multiply eqn (2) by 2 and then add to eqn 1 to get
x1 + x2 +x3 + 2(y1 + y2 + y3) = 19
x1 + 2y1 + x2 +x3 + 2(y2 + y3) = 19  now substitute value of first two terms from eqn (3)
this becomes   x2 +x3 + 2(y2 + y3) = 0   - (4)
now since, (x2,y2) and (x3,y3) lie on the line y = 2x - 10 the above we can write the qabove equation as
x2 +x3 + 2(2x2 + 2x3 -20) = 0
5(x2 +x3) = 40
x2 + x3 = 8      now we know what is x2 + x3 we can get value of y2 + y3 frm eqn (4) we get y2 + y3 = -4
now put value of x2 + x3 and y2 + y3 in eqn (1) and (2) resp to get x1 = 1 and y1 = 9
thus required point is (1,9)..... thats our Answer!!!!!
```
4 years ago

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