Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by ${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0 is$ $\frac{| c |}{\sqrt{(a - b)^{2} + 4 h^{2}}}$

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Detailed Solution

Given equation
$S = {ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$
Let the lines represented by S = 0 be
$l_{1} x + m_{1} y + n_{1} = 0 . . . (1)$
$and l_{2} x + m_{2} y + n_{2} = 0 \dots (2)$
$\begin{array}{l} ∴ {ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c \\ = (l_{1} x + m_{1} y + n_{1}) (l_{2} x + m_{2} y + n_{2}) \end{array}$
Comparing the coefficients of like terms we get
$\begin{array}{l} l_{1} l_{2} = a; m_{1} m_{2} = b; n_{1} n_{2} = c \\ l_{1} m_{2} + m_{1} l_{2} = 2 h, l_{1} n_{2} + n_{1} l_{2} = 2 g, m_{1} n_{2} + m_{2} n_{1} = 2 f \end{array}$
Perpendicular distance from origin to the line $(1) is \frac{|n_{1}|}{\sqrt{l_{1}^{2} + m_{1}^{2}}}$
Perpendicular distance from origin to the line (2) is $\frac{|n_{2}|}{\sqrt{l_{2}^{2} + m_{2}^{2}}}$
Product of the perpendicular distance from (0, 0) to S = 0 is
$\begin{array}{l} \frac{|n_{1}|}{\sqrt{l_{1}^{2} + m_{1}^{2}}} \frac{|n_{2}|}{\sqrt{l_{2}^{2} + m_{2}^{2}}} \\ = \frac{|n_{1} n_{2}|}{\sqrt{l_{1}^{2} l_{2}^{2} + m_{1}^{2} l_{2}^{2} + l_{1}^{2} m_{2}^{2} + m_{1}^{2} m_{2}^{2}}} \\ = \frac{| c |}{\sqrt{{(l_{1} l_{2} - m_{1} m_{2})}^{2} + 2 l_{1} l_{2} m_{1} m_{2} + {(l_{1} m_{2} + m_{1} l_{2})}^{2} - 2 l_{1} l_{2} m_{1} m_{2}}} \\ = \frac{| c |}{\sqrt{(a - b)^{2} + (2 h)^{2}}} = \frac{| c |}{\sqrt{(a - b)^{2} + 4 h^{2}}} \end{array}$
Hence proved