The distance between the two lines represented by the equation $9 x^{2} - 24 x y + 16 y^{2} - 12 x + 16 y - 12 = 0$ is __ units

\frac{8}{5}

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\frac{6}{5}

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\frac{11}{5}

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\frac{9}{5}

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Solution

The correct option is A $\frac{8}{5}$
$9 x^{2} - 24 x y + 16 y^{2} - 12 x + 16 y - 12 = 0 \Rightarrow (3 x - 4 y)^{2} - 4 (3 x - 4 y) - 12 = 0$
$\Rightarrow (3 x - 4 y + 2) (3 x - 4 y - 6) = 0$
So, the lines will be $3 x - 4 y + 2 = 0; 3 x - 4 y - 6 = 0$
Hence, distance between lines is
$\frac{| - 6 - 2 |}{5} = \frac{8}{5} units$

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