Median of triangle

**arze** · May 30th 2010, 11:04 PM

One of the medians of the triangle formed by the line pair $ax^2+2hxy+by^2=0$ and the line $px+qy=r$ 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, $y=\frac{r}{q}$
I found the sum of the points of intersection:
$x=\frac{r-qy}{p}$
$a(\frac{r-qy}{p})^2+2hy(\frac{r-qy}{p})+by^2=0$
$y^2(aq^2+bp^2-2hpq)+y(2r(hp-aq))+ar^2=0$
then
$-\frac{1}{2}\times\frac{2r(hp-aq)}{aq^2+bp^2-2hpq}=\frac{r}{q}$
works out to hq-bp=0
I can't see where i have gone wrong.
Thanks

**sa-ri-ga-ma** · May 31st 2010, 02:15 AM

Your work out is correct.

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