If 3a + 5b + 6c = 0,then the family of lines | vedclass

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(D) Given the equation of the family of lines: $ax + by + c = 0$. From the condition $3a + 5b + 6c = 0$,we can write $c = -\frac{3a + 5b}{6}$. Substit

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$\left(\frac{1}{2}, \frac{5}{6}\right)$

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(D) Given the equation of the family of lines: $ax + by + c = 0$. From the condition $3a + 5b + 6c = 0$,we can write $c = -\frac{3a + 5b}{6}$. Substituting this into the line equation: $ax + by - \frac{3a + 5b}{6} = 0$. Multiplying by $6$: $6ax + 6by - 3a - 5b = 0$. Rearranging the terms: $a(6x - 3) + b(6y - 5) = 0$. For this to hold for all $a$ and $b$,we must have $6x - 3 = 0$ and $6y - 5 = 0$. Solving these gives $x = \frac{3}{6} = \frac{1}{2}$ and $y = \frac{5}{6}$. Thus,the fixed point is $\left(\frac{1}{2}, \frac{5}{6}\right)$.

If $3a + 5b + 6c = 0$,then the family of lines $ax + by + c = 0$ passes through the fixed point:

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