Show that the area of the triangle formed by the lines $a x^{2} + 2 h x y + b y^{2} = 0$ and $l x + m y + n = 0$ is $\frac{n^{2} \sqrt{h^{2} - a b}}{∣ ∣ a m^{2} - 2 h l m + b l^{2} ∣ ∣}$

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Solution

Let $y = m_{1} x - - - - - (1)$
$y = m_{2} x - - - - - (2)$
be the lines represented by $a x^{2} + 2 b x y + b y^{2}$
Then $m_{1} + m_{2} = - 2 h / b$ and $m_{1} m_{2} = a / b - - - - - (3)$
The point of intersection of $y . m_{1} x$ and $x + m y .1$
is $[1 / l + m m_{1}]$ . $[m_{1} / 1 + m m_{1}]$ and that of
$y = m_{2} x$ and $l x + m y = 1$ is $[1 / l + m m_{2}] . [m_{2} / 1 + m m_{2}]$
Thus the three vertices are $(0, 0), (1 / l + m m_{1})$
$(m_{1} / l + m m_{2})$ and $(1 / 1 + m m_{2}), m_{2} / 1 + m . m_{1},$
so , the area of the triangle is $[1 / 2. (x_{1} y_{2} - x_{2} y_{1})]$
$. \frac{1}{2} [(1 / 1 + m m_{1}) (m_{2} / 1 + m m_{2})] + [(1 / 1 + m m_{2}) (m_{1} / 1 + m m_{1})]$
$: (m_{1} + m_{2}) / 2 (1 + m m_{1}) (1 + m m_{2})$
$. (m_{1} + m_{2}) / 2. [1 + m (m_{1} + m_{2}) + m^{2} . m_{1} m_{2} - - - - - (4)]$
Substituting $m_{1} + m_{2} - 2 h / b$ and
$m_{1} . m_{2} + a / b$ [from $(3)$ ]
Simplifying we get the area as
$= \frac{n^{2} . \sqrt{h^{2} . a b}}{∣ ∣ l a m^{2} - 2 h l m + b l^{2} ∣ ∣}$

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