A triangle has lines y=mx1 and y=mx2 as two of its sides where m1 and m2 are the roots of bx^2 + 2hx + a = 0. If P(a,b) is the orthocentre of the triangle then prove that the equation of its third side is (a+b)(ax+by) = ab(a+b-2h)

The given lines y=m1x and y=m2x intersect at O(0,0). Thus one vertex of the triangle be at origin.
Therefore let OAB be the triangle and OA, OB be the lines.

y=m1x..................(1)
and
y=m2x.................(2)  respectively
Let the equation of third side be
y=mx+c........................(3)
Given H(a,b) be the ortho centre of triangle OAB, it means OH is perpendicular to AB
So, (b/a)m=-1
or, m=-a/b
Solving (1) and (3) and (1) and (2), the coordinates of A are
$$\begin{gathered} \left(\frac{c}{m1-m},\frac{cm2}{m2-m}\right) \\ and of B are \\ \left(\frac{c}{m2-m},\frac{cm2}{m2-m}\right) \end{gathered}$$
$$\begin{gathered} Now equation of line through A and perpendicular to OB s \\ y-\frac{cm1}{m1-m}=\frac{-1}{m2}\left(x-\frac{c}{m1-m}\right) \\ or, y=\frac{-x}{m2}+\frac{c(m1m2+1)}{m2(m1-m)}........................\left(4\right) \\ Similarly equaion of line through B and perpendicular to OA is \\ y=\frac{-x}{m1}+\frac{c(m1m2+1)}{m1(m1-m)}........................\left(5\right) \\ The point of intersection of \left(4\right) and \left(5\right) is the ortho-centre H(a,b) \\ So, subtracting \left(5\right) and \left(4\right), we get \\ x=a=\frac{cm(m1m2+1)}{(m1-m)(m2-m)} \\ or, c=\frac{-[m1m2-m(m1+m2)+m^2]a}{m(m1m2+1)} \\ Since m1 and m2 are roots of the equation \\ b{x}^2+2hx+a=0 \\ So, m1+m2=-2h/b and m1m12=1/b \\ So, we hahe \\ c=\frac{-[a/b+2hm/b+m^2]a}{m(a/b+1)} \\ c=\frac{(a+2hm+b{m}^2)a}{m(a+b)} \\ Putting value of c in equation \left(3\right) we have \\ y=mx-\frac{(a+2hm+b{m}^2)a}{m(a+b)} \\ or, y=-\frac{a}{b}x-\frac{(a-2ha/b+b{\displaystyle \frac{a^2}{b^2}})a}{(-a/b)(a+b)} \\ or by solving we have, \\ (ax+by)(a+b)=ab(a+b-2h) \end{gathered}$$
Hence proved.