The equation of the pair of lines passing thr | vedclass

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

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

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(A) Given the pair of lines: $4xy + 2x + 6y + 3 = 0$. Factoring the expression: $2x(2y + 1) + 3(2y + 1) = 0$,which gives $(2x + 3)(2y + 1) = 0$. Thus,the 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,and a line perpendicular to $y = -\frac{1}{2}$ is a vertical line. The line passing through $(2, 1)$ perpendicular to $x = -\frac{3}{2}$ is $y = 1$. The line passing through $(2, 1)$ perpendicular to $y = -\frac{1}{2}$ is $x = 2$. Combining these equations: $(y - 1)(x - 2) = 0$. Expanding this gives $xy - 2y - x + 2 = 0$,or $xy - x - 2y + 2 = 0$.

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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