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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).
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