let ABC be triangle having orthocenter circumcenter at (9,5) (0,0) respectively if equation of side BC is2x-y=10 then find possible coordinte of vertex A.

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Solution

$Let the coordinates of vertices A, B and C be, (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) As we know that Centroid G, divides orthocentre O and cicumcentre C' in the ratio 2 : 1 so, coordinates of centroid G is = (\frac{2 \times 0 + 9 \times 1}{2 + 1}, \frac{2 \times 0 + 5 \times 1}{2 + 1}) = (3, \frac{5}{3}) Now let the perpendicular bisector of line BC be D having coordinates (x, y) then y = 2 x - 10 (since D lies on line BC) . So, D = (x, 2 x - 10) Also C' D and BC are perpendicular which implies - Slop of C' D \times slop of BC = - 1 Here cicumcentre is C' (0, 0) and D = (x, 2 x - 10) so, \frac{2 x - 10 - 0}{x - 0} \times 2 = - 1 from here, x = \frac{5}{2} so D = (\frac{5}{2}, - 5) Now this is the perpendicular bisector of BC so, \frac{x_{2} + x_{3}}{2} = \frac{5}{2} and \frac{y_{2} + y_{3}}{2} = - 5 x_{2} + x_{3} = 5 and y_{2} + y_{3} = - 10 . . . . . . (1) Now centroid G = (3, \frac{5}{3}) = [(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})] so x_{1} + x_{2} + x_{3} = 9 and y_{1} + y_{2} + y_{3} = 5 . . . . . . . . (2) From equation (1), and equation (2) we get, x_{1} = 4 and y_{1} = 15 So A = (4, 15)$ `

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