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.
MY TRY: Let coordinates of vertices A, B and C be $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ Centroid comes out as $(3,5/3)$.
Line joining the othocenter and $(x_1,y_1)$ is perpendicular to the base $2x - y = 10$, so slope of the line will be $-1/2$
We know it will pass through $(9,5)$ i.e the orthocenter so, using point slope form, we have equation of line as $2y + x = 19$
What's next? Solving for all 6 variables seems somewhat tedious..