The area (in sq units) of the quadrilateral f | vedclass

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(A) The given lines are of the form $a_1x + b_1y + c_1 = 0$,$a_1x - b_1y + c_1 = 0$,$a_1x + b_1y + c_2 = 0$,and $a_1x - b_1y + c_2 = 0$. The area of t

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$12$

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(A) The given lines are of the form $a_1x + b_1y + c_1 = 0$,$a_1x - b_1y + c_1 = 0$,$a_1x + b_1y + c_2 = 0$,and $a_1x - b_1y + c_2 = 0$. The area of the quadrilateral formed by these lines is given by the formula $	ext{Area} = \left| \frac{(c_1 - c_2)^2}{a_1b_1} ight|$. Here,$a_1 = 2$,$b_1 = 3$,$c_1 = 6$,and $c_2 = -6$. Substituting these values: $	ext{Area} = \left| \frac{(6 - (-6))^2}{2 	imes 3} ight| = \left| \frac{(12)^2}{6} ight| = \frac{144}{6} = 24$. Wait,re-evaluating the formula for lines $a_1x \pm b_1y + c_1 = 0$ and $a_1x \pm b_1y + c_2 = 0$: The lines are $L_1: 2x + 3y + 6 = 0$,$L_2: 2x + 3y - 6 = 0$,$L_3: 2x - 3y + 6 = 0$,$L_4: 2x - 3y - 6 = 0$. The distance between $L_1$ and $L_2$ is $d_1 = \frac{|6 - (-6)|}{\sqrt{2^2 + 3^2}} = \frac{12}{\sqrt{13}}$. The distance between $L_3$ and $L_4$ is $d_2 = \frac{|6 - (-6)|}{\sqrt{2^2 + (-3)^2}} = \frac{12}{\sqrt{13}}$. The angle $	heta$ between the lines $2x + 3y = 0$ and $2x - 3y = 0$ is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1m_2} ight| = \left| \frac{-2/3 - 2/3}{1 + (-2/3)(2/3)} ight| = \left| \frac{-4/3}{1 - 4/9} ight| = \left| \frac{-4/3}{5/9} ight| = \frac{12}{5}$. The area of the parallelogram is $\frac{d_1 d_2}{\sin 	heta}$. Since $	an 	heta = 12/5$,$\sin 	heta = 12/13$. $	ext{Area} = \frac{(12/\sqrt{13}) 	imes (12/\sqrt{13})}{12/13} = \frac{144/13}{12/13} = 12$ sq units.

The area (in sq units) of the quadrilateral formed by the lines $2x + 3y + 6 = 0$,$2x - 3y + 6 = 0$,$2x + 3y - 6 = 0$,and $2x - 3y - 6 = 0$ is

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