The straight lines 4ax + 3by + c = 0,where a | vedclass

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(B) Given the equation of the lines is $4ax + 3by + c = 0$ with the condition $a + b + c = 0$.From the condition,we can write $c = -(a + b)$.Substitut

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$(1/4, 1/3)$

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(B) Given the equation of the lines is $4ax + 3by + c = 0$ with the condition $a + b + c = 0$.From the condition,we can write $c = -(a + b)$.Substituting this into the line equation:$4ax + 3by - (a + b) = 0$Rearranging the terms to group $a$ and $b$:$a(4x - 1) + b(3y - 1) = 0$For this equation to hold for all values of $a$ and $b$,the coefficients must be zero:$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 straight lines $4ax + 3by + c = 0$,where $a + b + c = 0$,will be concurrent at the point:

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