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Question

$The distance between the lines given by (x + 7 y)^{2} - 42 = 4 \sqrt{2} (x + 7 y) is$

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Solution

$T h e c o m b i n e d e q u a t i o n o f t h e g i v e n l i n e s a r e {(x + 7 y)}^{2} - 4 \sqrt{2} (x + 7 y) - 42 = 0 S p l i t t i n g a s t w o e q u a t i o n s {(x + 7 y)}^{2} - 7 \sqrt{2} (x + 7 y) + 3 \sqrt{2} (x + 7 y) - 42 = 0 (x + 7 y) (x + 7 y - 7 \sqrt{2}) + 3 \sqrt{2} (x + 7 y - 7 \sqrt{2}) = 0 (x + 7 y - 7 \sqrt{2}) (x + 7 y + 3 \sqrt{2}) = 0$
$The given lines are x + 7 y = 7 \sqrt{2} and x + 7 y = - 3 \sqrt{2} .$
$Distance between lines = |\frac{C_{2} - C_{1}}{\sqrt{a^{2} + b^{2}}}| = \frac{| 10 \sqrt{2} |}{5 \sqrt{2}} = 2$

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If the lines represented by the equation $3 y^{2} - x^{2} + 2 \sqrt{3} x - 3 = 0$ are rotated about the point $(\sqrt{3}, 0)$ through an angle of $15^{0},$ one in clockwise direction and the other in anticlockwise direction, so that they become perpendicular, then the equation of the pair of lines in the new position is

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