???

one more doubt

find the area of triangle whose sides (lines are given by )

1) ax²+2hxy+by²=0
2) lx+my=1

The given pair of lines
ax^2+2hxy+by^2=0 ---- (1)
and the given line
lx+my=1 ----- (2)
Since the lines in (1) meet at the origin O, it is one of the vertices. Let P(x₁,y₁) and Q(x₂, y₂) be the other two vertices determined by intersections of (2) with lines given by (1).
Since P and Q lie on (2), hence
y_1=\dfrac{1-l x_1}{m} and y_2=\dfrac{1-l x_2}{m}
Hence y_1-y_2=\dfrac{l}{m}(x_2-x_1)
Hence,
PQ=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\dfrac{\sqrt{l^2+m^2}}{m}\sqrt{(x_1-x_2)^2}
i.e.
PQ=\dfrac{\sqrt{l^2+m^2}}{m}\sqrt{(x_1+x_2)^2-4x_1x_2}
Now, x_1,2 are the roots of the equation obtained by eliminating y in (1) and (2) i.e. the equation
(am^2-2hlm+bl^2)x^2+2(hm-bl)x+b=0
Hence,
x_1+x_2=-\dfrac{2(hm-bl)}{am^2-2hlm+bl^2}
and
x_1x_2=\dfrac{b}{am^2-2hlm+bl^2}
So
(x_1+x_2)^2-4x_1x_2=\dfrac{4(hm-bl)^2-4b(am^2-2hlm+bl^2)}{(am^2-2hlm+bl^2)^2}
\Rightarrow\ (x_1+x_2)^2-4x_1x_2=\dfrac{4m^2(h^2-ab)}{(am^2-2hlm+bl^2)^2}
So PQ becomes
PQ=\dfrac{2\sqrt{(l^2+m^2)(h^2-ab)}}{am^2-2hlm+bl^2}
On the other hand the perpendicular dropped from O to PQ is
d = \dfrac{1}{\sqrt{l^2+m^2}}
Hence the required area is
\dfrac{1}{2}PQ\times d = \dfrac{\sqrt{h^2-ab}}{am^2-2hlm+bl^2}

Same here Sir i was also thinking on the same procedure..but i dint proceed further coz it got too long due to so many calculations...It's so nice of u to post the entire solution..

Kaymant Sir
[240]

wat a nice solution .........
U r gr8 KAYMANT sir.
same thng that GOVIND wants to say abt u........[1][1]