Median of triangle

Jul 2009
338
14
Singapore
One of the medians of the triangle formed by the line pair \(\displaystyle ax^2+2hxy+by^2=0\) and the line \(\displaystyle px+qy=r\) lies along the y-axis. If neither a nor r is zero, prove that bp+hq=0.
The line pair has its vertex at (0,0), so that is the vertex of the median on the y-axis. Then the midpoint of the intersection of the line pair and the line would be the other point on the median and triangle.
When x=0, on the line, \(\displaystyle y=\frac{r}{q}\)
I found the sum of the points of intersection:
\(\displaystyle x=\frac{r-qy}{p}\)
\(\displaystyle a(\frac{r-qy}{p})^2+2hy(\frac{r-qy}{p})+by^2=0\)
\(\displaystyle y^2(aq^2+bp^2-2hpq)+y(2r(hp-aq))+ar^2=0\)
then
\(\displaystyle -\frac{1}{2}\times\frac{2r(hp-aq)}{aq^2+bp^2-2hpq}=\frac{r}{q}\)
works out to hq-bp=0
I can't see where i have gone wrong.
Thanks