If the lines $a x^{2} + 2 h x y + b y^{2} = 0$ be two sides of a parallelogram and the line $l x + m y = 1$ be one of its diagonal, then $y (b l - h m) - x (a m - h l) = ?$ .

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Solution

Consider the diagram shown below.

Equation of $O C \Rightarrow y = m_{1} x$
Equation of $O A \Rightarrow y = m_{2} x$

Given:
Equation of $A C \Rightarrow l x + m y = 1$
Equation of the sides $\Rightarrow a x^{2} + 2 h x y + b y^{2} = 0$

Therefore,
$m_{1} + m_{2} = - \frac{2 h}{b}$
$m_{1} m_{2} = \frac{a}{b}$

$C$ is the intersection point of $A C$ and $O C$ . Therefore,
$\Rightarrow l x + m (m_{1} x) = 1$
$\Rightarrow x = \frac{1}{l + m m_{1}}$
$\Rightarrow y = m_{1} x = \frac{m_{1}}{l + m m_{1}}$

Also, $A$ is the intersection point of $A C$ and $O A$ . Therefore,
$\Rightarrow l x + m (m_{2} x) = 1$
$\Rightarrow x = \frac{1}{l + m m_{2}}$
$\Rightarrow y = m_{2} x = \frac{m_{2}}{l + m m_{2}}$

Let $P (α, β)$ be the mid-point of $A C$ . Then,
$α = \frac{1}{2} (\frac{1}{l + m m_{1}} + \frac{1}{l + m m_{2}})$ ……. (1)
$β = \frac{1}{2} (\frac{m_{1}}{l + m m_{1}} + \frac{m_{2}}{l + m m_{2}})$ …… (2)

From equations (1) and (2), we have
$\Rightarrow \frac{α}{β} = \frac{2 l + m (m_{1} + m_{2})}{m_{1} (l + m m_{2}) + m_{2} (l + m m_{1})}$
$\Rightarrow \frac{α}{β} = \frac{2 l + m (- \frac{2 h}{b})}{l (- \frac{2 h}{b}) + 2 m (\frac{a}{b})}$
$\Rightarrow \frac{α}{β} = \frac{b l - m h}{a m - h l}$

We know that, diagonals of a parallelogram bisect each other. So, $P$ will also lie on $O B$ . Two points on $O B$ are $(0, 0)$ and $P (α, β)$ . Therefore, equation of $O B$ is,
$\Rightarrow y = \frac{β}{α} x$
$\Rightarrow y = \frac{a m - h l}{b l - h m} x$
$\Rightarrow y (b l - h m) = x (a m - h l)$
$∴ y (b l - h m) - x (a m - h l) = 0$
Hence, this is the required answer.

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