P.T the area of the triangle formed by the pair of linesax^2 + 2hxy+by^2=0 and lx+my+n=0 is (n^2Root(h2-ab) / |am^2=2hlm=bl^2

Let y=m1 x ………..(1) and y=m2 x …………(2) be the lines represented by ax^2 +2hxy+by^2= 0. Then m1 + m2 = -2h/b and m1 m2 = a/b…….(3)

Now the point of intersection of y =m1 x and lx +my= 1 is (1/(l+mm1), m1/(1+mm1))

and

that of y=m2 x and lx+my=1 is (1/(l+m m2), m2/(1+m m2)).

Thus the three vertices are (0,0), (1/(l+mm1), m1/(1+mm1)) and (1/(l+m m2), m2/(1+m m2)).

So, the area of the triangle is (1/2)(x1 y2 -x2 y1) = (1/2)[{1/(l+mm1)}{m2/(1+mm2)}+ {1/(l+mm2)}{m1/(1+mm1)}

= (m1 + m2)/2(1+mm1)(1+mm2)= (m1+m2)/2{1+m(m1+m2)+m^2 m1m2}………..(4).

Substituting m1 + m2 = -2h/b and m1 m2 = a/b from (3) in (4) and simplifying we get the area as √(h^2-ab)/(am^2-2hlm+bl^2).