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..

  • $\begingroup$ Hint: If H is the orthocenter and OF is the perpendicular from O onto BC, then AH = 2OF. Do you realize what do now? You won't even need to calculate the centroid. :D $\endgroup$ – Sawarnik Sep 15 '16 at 21:39

You have now that $2y_1+x_1=19$, i.e. $x_1=19-2y_1$.

Also, note that you can have $2x_2-y_2=10$ and $2x_3-y_3=10$.

From $OB=OC$ where $O(0,0)$ with $x_2\not=x_3$, $$x_2^2+(2x_2-10)^2=x_3^2+(2x_3-10)^2\implies x_3+x_2=8$$

Since you know that the centroid is $(3,5/3)$, you can have $3=\frac{19-2y_1+x_2+x_3}{3}\implies y_1=9$.

Hence, $A(1,9)$.


You have already found out the equation of the altitude passing through the vertex A. The foot of perpendicular from origin to the line BC will give you the coordinates of the midpoint of BC .

Using the midpoint and the centroid you have the equation of the median through A.

I think we can find the coordinates of A now.

EDIT: On solving, the centroid and vertex A seem to be lying on opposite sides of BC .


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