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Thread: Median of triangle

  1. #1
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    Median of triangle

    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
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  2. #2
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    Your work out is correct.
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