If $a^{2} + 9 b^{2} - 4 c^{2} = 6 a b$ , then the family of line $a x + b y + c = 0$ are concurrent at:

(\frac{1}{2}, \frac{3}{2}) (- \frac{1}{3}, \frac{- 3}{2})

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(- \frac{1}{2}, \frac{3}{2}) (\frac{1}{2}, \frac{- 3}{2})

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(- \frac{1}{7} - \frac{3}{2}) (- \frac{1}{5}, \frac{- 3}{2})

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(\frac{- 1}{7}, \frac{3}{2}) (- \frac{1}{2}, \frac{- 3}{2})

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Solution

The correct option is B $(- \frac{1}{2}, \frac{3}{2}) (\frac{1}{2}, \frac{- 3}{2})$
Given that,
$a^{2} + 9 b^{2} - 4 c^{2} = 6 a b$

$⟹ a^{2} + 9 b^{2} - 6 a b - 4 c^{2} = 0$

$⟹ (a - 3 b)^{2} - (2 c)^{2} = 0$

$⟹ (a - 3 b - 2 c) (a - 3 b + 2 c) = 0$

$⟹ (\frac{- 1}{2} a + \frac{3}{2} b + c) (\frac{1}{2} a + \frac{- 3}{2} b + c) = 0$

$∴$ The family of line $a x + b y + c = 0$ are concurrent at $(\frac{- 1}{2}, \frac{3}{2}), (\frac{1}{2}, \frac{- 3}{2})$

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Similar questions

a^{2} + 9 b^{2} - 4 c^{2} = 6 a b

a x + b y + c = 0

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m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

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If the three lines $y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and $y = m_{3} x + c_{3}$ are concurrent then show that,

$m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0$

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