the distance between the parallel lines 9x2-24xy+16y2-12x+16y-12=0 is

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Solution

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$9 x^{2} - 24 x y + 16 y^{2} - 12 x + 16 y - 12 = 0 (3 x)^{2} - 2.3 x . 4 y + (4 y)^{2} - 4 (3 x - 4 y) - 12 = 0 (3 x - 4 y)^{2} - 4 (3 x - 4 y) - 12 = 0 (3 x - 4 y)^{2} - 6 (3 x - 4 y) + 2 (3 x - 4 y) - 12 = 0 (3 x - 4 y) (3 x - 4 y - 6) + 2 (3 x - 4 y - 6) = 0 (3 x - 4 y - 6) (3 x - 4 y + 2) = 0 s o l i n e s a r e 3 x - 4 y - 6 = 0 3 x - 4 y + 2 = 0 d i s \tan c e = \frac{|6 - (- 2)|}{\sqrt{3^{2} + 4^{2}}} = \frac{8}{5}$

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9 x^{2} - 24 x y + 16 y^{2} - 12 x + 16 y - 12 = 0

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