Consider the set of all lines px + qy + r = 0 | vedclass

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(A) Given the equation of the line $px + qy + r = 0$ and the condition $3p + 2q + 4r = 0$. From the condition,we can write $r = -\frac{3p + 2q}{4}$. S

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The lines are concurrent at the point $\left( \frac{3}{4}, \frac{1}{2} \right)$

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(A) Given the equation of the line $px + qy + r = 0$ and the condition $3p + 2q + 4r = 0$. From the condition,we can write $r = -\frac{3p + 2q}{4}$. Substituting this into the line equation: $px + qy - \frac{3p + 2q}{4} = 0$ $4px + 4qy - 3p - 2q = 0$ Rearranging the terms to group $p$ and $q$: $p(4x - 3) + q(4y - 2) = 0$ For this to hold for all $p$ and $q$,the coefficients must be zero: $4x - 3 = 0 \implies x = \frac{3}{4}$ $4y - 2 = 0 \implies y = \frac{2}{4} = \frac{1}{2}$ Thus,all lines pass through the fixed point $\left( \frac{3}{4}, \frac{1}{2} \right)$,meaning they are concurrent at this point.

Consider the set of all lines $px + qy + r = 0$ such that $3p + 2q + 4r = 0$. Which one of the following statements is true?

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