The area of the parallelogram formed by the l | vedclass

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(A) The area of a parallelogram formed by two pairs of parallel lines $L_1: a_1x + b_1y + c_1 = 0, L_2: a_1x + b_1y + c_2 = 0$ and $M_1: a_2x + b_2y +

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$|p - q| |r - s| |\csc(\alpha - \beta)|$

Vedclass · Answer

(A) The area of a parallelogram formed by two pairs of parallel lines $L_1: a_1x + b_1y + c_1 = 0, L_2: a_1x + b_1y + c_2 = 0$ and $M_1: a_2x + b_2y + d_1 = 0, M_2: a_2x + b_2y + d_2 = 0$ is given by $	ext{Area} = \frac{|c_1 - c_2| |d_1 - d_2|}{|a_1b_2 - a_2b_1|}$.Here,the lines are:$x \cos \alpha + y \sin \alpha - p = 0$$x \cos \alpha + y \sin \alpha - q = 0$$x \cos \beta + y \sin \beta - r = 0$$x \cos \beta + y \sin \beta - s = 0$Distance between the first pair of parallel lines is $d_1 = |p - q|$.Distance between the second pair of parallel lines is $d_2 = |r - s|$.The angle $	heta$ between the lines is $\alpha - \beta$.The denominator is $|(\cos \alpha)(\sin \beta) - (\cos \beta)(\sin \alpha)| = |\sin(\beta - \alpha)| = |\sin(\alpha - \beta)|$.Thus,$	ext{Area} = \frac{|p - q| |r - s|}{|\sin(\alpha - \beta)|} = |p - q| |r - s| |\csc(\alpha - \beta)|$.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

The area of the parallelogram formed by the lines $x \cos \alpha + y \sin \alpha = p$,$x \cos \alpha + y \sin \alpha = q$,$x \cos \beta + y \sin \beta = r$,and $x \cos \beta + y \sin \beta = s$ is:

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