Question

A triangle has the lines ax2+2hxy+by2=0 for two of its sides and the point

Answer 1

Answer:

triangle has the lines X square + 2 HD y + fever 242 of a child.

Answer 2

Answer:

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