The linear factor(s) of the equation $9 x^{2} - 24 x y + 16 y^{2} - 12 x + 16 y - 12 = 0$ is/are

3 x - 2 y + 2 = 0

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3 x - 2 y + 6 = 0

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3 x - 4 y - 6 = 0

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3 x - 4 y + 2 = 0

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Solution

The correct option is D $3 x - 4 y + 2 = 0$
$9 x^{2} - 24 x y + 16 y^{2} - 12 x + 16 y - 12 = 0$
Writing equation in terms of $y,$ we get
$16 y^{2} + (- 24 x + 16) y + (9 x^{2} - 12 x - 12) = 0$
Now, using quadratic formula, we have
$y = \frac{24 x - 16 \pm \sqrt{(- 24 x + 16)^{2} - 4 \cdot 16 \cdot (9 x^{2} - 12 x - 12)}}{2 \cdot 16} \Rightarrow y = \frac{24 x - 16 \pm \sqrt{64 (3 x - 2)^{2} - 64 (9 x^{2} - 12 x - 12)}}{2 \cdot 16} \Rightarrow y = \frac{24 x - 16 \pm 8 \sqrt{(9 x^{2} + 4 - 12 x) - (9 x^{2} - 12 x - 12)}}{2 \cdot 16} \Rightarrow y = \frac{3 x - 2 \pm 4}{4} \Rightarrow y = \frac{3 x + 2}{4} or y = \frac{3 x - 6}{4}$
$\Rightarrow 3 x - 4 y + 2 = 0 or 3 x - 4 y - 6 = 0$

Alternate Solution:
$9 x^{2} - 24 x y + 16 y^{2} - 12 x + 16 y - 12 = 0 \Rightarrow (9 x^{2} - 24 x y + 16 y^{2}) - 4 (3 x - 4 y) - 12 = 0 \Rightarrow (3 x - 4 y)^{2} - 4 (3 x - 4 y) - 12 = 0$
$\Rightarrow (3 x - 4 y + 2) (3 x - 4 y - 6) = 0$
$\Rightarrow 3 x - 4 y + 2 = 0 or 3 x - 4 y - 6 = 0$

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