At which point are the lines ax + by + c = 0 | vedclass

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(B) The given equation is $ax + by + c = 0$ and the condition is $3a + 2b + 4c = 0$.From the second equation,we have $c = -\frac{3a + 2b}{4}$.Substitu

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$(3/4, 1/2)$

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(B) The given equation is $ax + by + c = 0$ and the condition is $3a + 2b + 4c = 0$.From the second equation,we have $c = -\frac{3a + 2b}{4}$.Substituting this into the first equation:$ax + by - \frac{3a + 2b}{4} = 0$Multiply by $4$ to clear the denominator:$4ax + 4by - 3a - 2b = 0$Rearranging terms to group $a$ and $b$:$a(4x - 3) + b(4y - 2) = 0$For this to be true for all $a$ and $b$,the coefficients must be zero:$4x - 3 = 0 \implies x = 3/4$$4y - 2 = 0 \implies y = 2/4 = 1/2$Thus,the point of concurrency is $(3/4, 1/2)$.

At which point are the lines $ax + by + c = 0$ and $3a + 2b + 4c = 0$ concurrent?

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