The family of straight lines 4ax + 3by + c = | vedclass

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(C) Given the equation of the family of lines: $4ax + 3by + c = 0$ $(1)$.Given the condition: $a + b + c = 0$, which implies $c = -a - b$ $(2)$.Substi

Vedclass · Accepted Answer

$(\frac{1}{4}, \frac{1}{3})$

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(C) Given the equation of the family of lines: $4ax + 3by + c = 0$ $(1)$.Given the condition: $a + b + c = 0$, which implies $c = -a - b$ $(2)$.Substitute $(2)$ into $(1)$:$4ax + 3by - a - b = 0$Rearrange the terms to group $a$ and $b$:$a(4x - 1) + b(3y - 1) = 0$For this equation to hold for all real constants $a$ and $b$, the coefficients of $a$ and $b$ must be zero independently:$4x - 1 = 0 \implies x = \frac{1}{4}$$3y - 1 = 0 \implies y = \frac{1}{3}$Thus, the lines are concurrent at the point $(\frac{1}{4}, \frac{1}{3})$.

The family of straight lines $4ax + 3by + c = 0$ such that $a + b + c = 0$ (where $a, b, c$ are real constants) are concurrent at the point...

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