Show that the product of the perpendicular distances from a point ( $α, β$ ) to the pair of straight lines ${ax}^{2} + 2 hxy + {by}^{2} = 0$ is $\frac{|{aα}^{2} + 2 hαβ + {bβ}^{2}|}{\sqrt{(a - b)^{2} + 4 h^{2}}}$

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Detailed Solution

$Let {ax}^{2} + 2 hxy + {by}^{2} = 0 represent the lines$
$l_{1} x + m_{1} y = 0 . . . (1) and l_{2} x + m_{2} y = 0 . . . (2)$
$Then l_{1} l_{2} = a, l_{1} m_{2} + l_{2} m_{1} = 2 h, m_{1} m_{2} = b$
$The lengths of perpendiculars from (α, β) to$
$the line (1) is p = \frac{|l_{1} α + m_{1} β|}{\sqrt{l_{1}^{2} + m_{1}^{2}}} and to the$
$\begin{array}{l} (2) is q = \frac{|l_{2} α + m_{2} β|}{\sqrt{{l_{2}}^{2} + {m_{2}}^{2}}} \\ pq = |\frac{l_{1} α + m_{1} β}{\sqrt{l_{1}^{2} + m_{1}^{2}}} \cdot| \cdot |\frac{l_{2} α + m_{2} β}{\sqrt{l_{2}^{2} + m_{2}^{2}}}| \\ = |\frac{l_{1} l_{2} α^{2} + (l_{1} m_{2} + l_{2} m_{1}) αβ + m_{1} m_{2} β^{2}}{\sqrt{l_{1}^{2} l_{2}^{2} + l_{1}^{2} m_{2}^{2} + l_{2}^{2} m_{1}^{2} + m_{1}^{2} m_{2}^{2}}}| = \frac{|{aα}^{2} + 2 hαβ + {bβ}^{2}|}{\sqrt{{(l_{1} l_{2} - m_{1} m_{2})}^{2} + 2 l_{1} l_{2} m_{1} m_{2} + {(l_{1} m_{2} + l_{2} m_{1})}^{2} - 2 l_{1} m_{2} l_{2} m_{1}}} \\ = \frac{|{aα}^{2} + 2 hαβ + {bβ}^{2}|}{\sqrt{(a - b)^{2} + 4 h^{2}}} \end{array}$