If 4a^2 + 9b^2 - c^2 + 12ab = 0, then the fam | vedclass

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(A) Given the equation $4a^2 + 12ab + 9b^2 - c^2 = 0$. This can be written as $(2a + 3b)^2 - c^2 = 0$. Using the identity $x^2 - y^2 = (x - y)(x + y)$

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$(2, 3)$ or $(-2, -3)$

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(A) Given the equation $4a^2 + 12ab + 9b^2 - c^2 = 0$. This can be written as $(2a + 3b)^2 - c^2 = 0$. Using the identity $x^2 - y^2 = (x - y)(x + y)$, we get $(2a + 3b - c)(2a + 3b + c) = 0$. This implies $2a + 3b - c = 0$ or $2a + 3b + c = 0$, which can be written as $c = \pm(2a + 3b)$. Substituting this into the line equation $ax + by + c = 0$, we get $ax + by \pm(2a + 3b) = 0$. Rearranging the terms, we get $a(x \pm 2) + b(y \pm 3) = 0$. For this to be true for all $a$ and $b$, the coefficients must be zero: $x \pm 2 = 0$ and $y \pm 3 = 0$. Thus, the lines are concurrent at $(2, 3)$ or $(-2, -3)$.

If $4a^2 + 9b^2 - c^2 + 12ab = 0$, then the family of straight lines $ax + by + c = 0$ is concurrent at

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