# There is a triangle having two sides $ax^2 +2hxy +by^2 = 0$ and orthocenter $(p, q)$ Find the other side.

There is a triangle having two sides $ax^2 +2hxy +by^2 = 0$ and orthocenter $(p, q)$. Find the equation of other side.

I first let the sides are $OM$: $y = m_1x$ and $ON:$ $y = m_2x$. Where $m_1m_2= \frac{-2h}{b}$ and $m_1 + m_2 = \frac{a}{b}$

Then I found the foot of the perpendicular of $(p, q)$ on $OM$. It was $$Q\left(\frac{p-m_1q}{m_1^2 + 1},m_1\frac{p-m_1q}{m_1^2 + 1}\right)$$

Now $Q$ is the midpoint of $O$ and $M$. Thus I got $M$ is $$\left(2\frac{p-m_1q}{m_1^2 + 1},2m_1\frac{p-m_1q}{m_1^2 + 1} \right)$$

Similarly, N is $$\left(2\frac{p-m_2q}{m_2^2 + 1} ,2m_2\frac{p-m_2q}{m_2^2 + 1} \right)$$

Using two point formula and making some manipulation, the equation of line $MN$ i.e. the third side became

$$\frac{(m_1^2 + 1)y- 2m_1p+2m_1^2q}{(m_1^2 + 1)x- 2p+2m_1q} =\frac{m_1m_2p}{(m_1+m_2)p+q}$$

From now where to go ?

Your method is incorrect. The orthocenter is simply the intersection of altitudes. It has nothing to do with midpoints.

Denote $H(p,q)$ as the orthocenter. The triangle is $\triangle OMN$. Then

• $HN$ is perpendicular to $OM$
• $HM$ is perpendicular to $ON$

What you should do

• Find one of the altitudes, let's say $HN$, by making the line through $H$ with slope $-1/m_2$ (from the slope of $OM$). $Q$ is indeed the intersection of $HN$ and $OM$, but it is not the midpoint of $OM$

• Find $N$ is by intersecting $HN$ with $ON$.

Find $M$ in the same way. Once you have $M$ and $N$, find the line through them. You result should contain $p,q,m_1,m_2$ as these are all known constants.

Alternatively, You can find $MN$ without finding both points. After finding $N$

• Find the line $OH$. This should be easy enough
• Find the line through $N$ and perpendicular to $OH$. This is $MN$

Put $A(0,0),B(x,0),C(?,?)$. The coordinates of the vertex $C$ are intersection of the perpendiculars to the altitudes passing by $A$ and $B$, so we have after easy calculation $C(p,\frac{p(x-p)}{q})$.

Now one has $$(AC)^2=p^2+\left(\frac{p(x-p)}{q}\right)^2\\(BC)^2=(p-x)^2+\left(\frac{p(x-p)}{q}\right)^2$$ The quotient $\dfrac{AC}{BC}$ is root of the given condition $ax^2+2hxy+by^2=0\iff a(\frac xy)^2+2h(\frac xy)+b=0$.

Consequently one has, putting $R=\dfrac{-2h\pm\sqrt{h^2-4ab}}{2a}$ (as data of problem)

$$\frac{(p-x)^2+\left(\frac{p(x-p)}{q}\right)^2}{p^2+\left(\frac{p(x-p)}{q}\right)^2}=R^2$$ This quadratic equation gives the value of $x$ from which we have calculated the three vertices of the triangle determining this way the three sides, $x$ itself being the required side.