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(A) The given equation is $9x^2 - 24xy + 16y^2 + 3x - 4y - 6 = 0$.This can be written as $(3x - 4y)^2 + (3x - 4y) - 6 = 0$.Let $u = 3x - 4y$. Then the

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(A) The given equation is $9x^2 - 24xy + 16y^2 + 3x - 4y - 6 = 0$.This can be written as $(3x - 4y)^2 + (3x - 4y) - 6 = 0$.Let $u = 3x - 4y$. Then the equation becomes $u^2 + u - 6 = 0$.Factoring the quadratic equation: $(u + 3)(u - 2) = 0$.So,the two lines are $3x - 4y + 3 = 0$ and $3x - 4y - 2 = 0$.The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \left|\frac{c_1 - c_2}{\sqrt{a^2 + b^2}}\right|$.Here,$a = 3, b = -4, c_1 = 3, c_2 = -2$.$d = \left|\frac{3 - (-2)}{\sqrt{3^2 + (-4)^2}}\right| = \left|\frac{5}{\sqrt{9 + 16}}\right| = \left|\frac{5}{\sqrt{25}}\right| = \frac{5}{5} = 1$.

The distance between the two lines represented by the pair of straight lines $9x^2 - 24xy + 16y^2 + 3x - 4y - 6 = 0$ is:

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