The area (in square units) of the quadrilater | vedclass

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(A) The given equations are:$l^2x^2 - m^2y^2 - n(lx + my) = 0 \implies (lx - my)(lx + my) - n(lx + my) = 0 \implies (lx + my)(lx - my - n) = 0$.Thus,t

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$\frac{n^2}{2|lm|}$

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(A) The given equations are:$l^2x^2 - m^2y^2 - n(lx + my) = 0 \implies (lx - my)(lx + my) - n(lx + my) = 0 \implies (lx + my)(lx - my - n) = 0$.Thus,the first pair of lines is $lx + my = 0$ and $lx - my = n$.$l^2x^2 - m^2y^2 + n(lx - my) = 0 \implies (lx - my)(lx + my) + n(lx - my) = 0 \implies (lx - my)(lx + my + n) = 0$.Thus,the second pair of lines is $lx - my = 0$ and $lx + my = -n$.The four lines are:$L_1: lx + my = 0$$L_2: lx - my = n$$L_3: lx - my = 0$$L_4: lx + my = -n$The area of the parallelogram formed by lines $a_1x + b_1y + c_1 = 0, a_1x + b_1y + c_2 = 0$ and $a_2x + b_2y + d_1 = 0, a_2x + b_2y + d_2 = 0$ is given by $\left| \frac{(c_1 - c_2)(d_1 - d_2)}{a_1b_2 - a_2b_1} \right|$.Here,$a_1=l, b_1=m, c_1=0, c_2=n$ (for $lx-my=0$ and $lx-my=n$) and $a_2=l, b_2=-m, d_1=0, d_2=-n$ (for $lx+my=0$ and $lx+my=-n$).Area = $\left| \frac{(0 - n)(0 - (-n))}{l(-m) - m(l)} \right| = \left| \frac{-n^2}{-2lm} \right| = \frac{n^2}{2|lm|}$.

The area (in square units) of the quadrilateral formed by the two pairs of lines $l^2x^2 - m^2y^2 - n(lx + my) = 0$ and $l^2x^2 - m^2y^2 + n(lx - my) = 0$ is

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