If the second degree equation S = ax² + 2hxy + by² + 2gx + 2fy + c = 0 in two variables x and y represents a pair of straight lines, then show that
$(i) abc + 2 fgh - {af}^{2} - {bg}^{2} - {ch}^{2} = 0 and (ii) h^{2} \geq ab, g^{2} \geq ac and f^{2} \geq bc$

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Detailed Solution

Let the equation S = 0 represents two lines $l_{1} x + m_{1} y + n_{1} = 0 and l_{2} x + m_{2} y + n_{2} = 0$
Then
$\begin{array}{l} {ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c \equiv (l_{1} x + m_{1} y + n_{1}) (l_{2} x + m_{2} y + n_{2}) \\ _{} \end{array}$
Equating the co-efficients of like terms on both sides
we get $l_{1} l_{2} = a, l_{1} m_{2} + l_{2} m_{1} = 2 h, m_{1} m_{2} = b$
$l_{1} n_{2} + l_{2} n_{1} = 2 g, m_{1} n_{2} + m_{2} n_{1} = 2 f, n_{1} n_{2} = c$
i) Consider the product (2h) (2g) (2f)
$\begin{array}{l} = (l_{1} m_{2} + l_{2} m_{1}) (l_{1} n_{2} + l_{2} n_{1}) (m_{1} n_{2} + m_{2} n_{1}) \\ = l_{1} l_{2} ({m_{1}}^{2} n_{2}^{2} + m_{2}^{2} n_{1}^{2}) + m_{1} m_{2} (l_{1}^{2} n_{2}^{2} + l_{2}^{2} n_{1}^{2}) \\ + n_{1} n_{2} (l_{1}^{2} {m_{2}}^{2} + l_{2} m_{1}^{2}) + 2 l_{1} l_{2} m_{1} m_{2} n_{1} n_{2} \\ = l_{1} l_{2} [{(m_{1} n_{2} + m_{2} n_{1})}^{2} - 2 m_{1} m_{2} n_{1} n_{2}] \\ + m_{1} m_{2} [{(l_{1} n_{2} + l_{2} n_{1})}^{2} - 2 l_{1} l_{2} n_{1} n_{2}] \\ + n_{1} n_{2} [{(l_{1} m_{2} + l_{2} m_{1})}^{2} - 2 l_{1} l_{2} m_{1} m_{2}] + 2 l_{1} l_{2} m_{1} m_{2} n_{1} n_{2} \\ = a (4 f^{2} - 2 bc) + b [4 g^{2} - 2 ac] + \\ c (4 h^{2} - 2 ab) + 2 abc \\ \Rightarrow 8 fgh = 4 [{af}^{2} + {bg}^{2} + {ch}^{2} - abc] \\ ∴ abc + 2 fgh - {af}^{2} - {bg}^{2} - {ch}^{2} = 0 \end{array}$
$\begin{array}{l} ii) h^{2} - ab = {(\frac{l_{1} m_{2} + l_{2} m_{1}}{2})}^{2} - l_{1} l_{2} m_{1} m_{2} \\ = \frac{{(l_{1} m_{2} + l_{2} m_{1})}^{2} - 4 l_{1} l_{2} m_{1} m_{2}}{4} \\ h^{2} - ab = \frac{{(l_{1} m_{2} - l_{2} m_{1})}^{2}}{4} \geq 0 \\ ∴ h^{2} \geq ab \end{array}$
Similarly we can prove $g^{2} \geq ac and f^{2} \geq bc$