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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.

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.

Grade:11

2 Answers

Priyansh Bajaj AskiitiansExpert-IITD
30 Points
10 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)

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All the best Aman!!!

Regards,
Askiitians Experts
Priyansh Bajaj

Ayoosh Kathuria
20 Points
8 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!!!!!

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