The equation of the pair of lines passing thr | vedclass

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(A) Given pair of lines: $4xy + 2x + 6y + 3 = 0$. Factorizing the expression: $2x(2y + 1) + 3(2y + 1) = 0$ $(2x + 3)(2y + 1) = 0$. Thus,the individual

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$xy - x - 2y + 2 = 0$

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(A) Given pair of lines: $4xy + 2x + 6y + 3 = 0$. Factorizing the expression: $2x(2y + 1) + 3(2y + 1) = 0$ $(2x + 3)(2y + 1) = 0$. Thus,the individual lines are $x = -\frac{3}{2}$ (a vertical line) and $y = -\frac{1}{2}$ (a horizontal line). $A$ line perpendicular to $x = -\frac{3}{2}$ is a horizontal line of the form $y = k$. Since it passes through $(2,1)$,$y = 1$ or $y - 1 = 0$. $A$ line perpendicular to $y = -\frac{1}{2}$ is a vertical line of the form $x = h$. Since it passes through $(2,1)$,$x = 2$ or $x - 2 = 0$. The combined equation of these two lines is $(y - 1)(x - 2) = 0$. Expanding this,we get $xy - 2y - x + 2 = 0$,which is $xy - x - 2y + 2 = 0$.

The equation of the pair of lines passing through the point $(2,1)$ and perpendicular to the pair of lines $4xy + 2x + 6y + 3 = 0$ is:

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