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 neitheranorris zero, prove thatbp+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 tohq-bp=0

I can't see where i have gone wrong.

Thanks