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

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(A) The given equations are $\lambda^2 x^2 - m^2 y^2 \mp n(\lambda x + m y) = 0$. Factoring these,we get: $(\lambda x - m y)(\lambda x + m y) - n(\lam

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$\frac{n^2}{2|\lambda m|}$

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(A) The given equations are $\lambda^2 x^2 - m^2 y^2 \mp n(\lambda x + m y) = 0$. Factoring these,we get: $(\lambda x - m y)(\lambda x + m y) - n(\lambda x + m y) = 0 \implies (\lambda x + m y)(\lambda x - m y - n) = 0$ $(\lambda x - m y)(\lambda x + m y) + n(\lambda x + m y) = 0 \implies (\lambda x + m y)(\lambda x - m y + n) = 0$ This represents two pairs of parallel lines: Pair $1$: $\lambda x + m y = 0$ and $\lambda x + m y + n = 0$ Pair $2$: $\lambda x - m y = 0$ and $\lambda x - m y - n = 0$ The area of the parallelogram formed by lines $a_1 x + b_1 y + c_1 = 0, a_1 x + b_1 y + c_2 = 0$ and $a_2 x + b_2 y + d_1 = 0, a_2 x + b_2 y + d_2 = 0$ is given by $\left| \frac{(c_1 - c_2)(d_1 - d_2)}{a_1 b_2 - a_2 b_1} \right|$. Here,$c_1 = 0, c_2 = n$ and $d_1 = 0, d_2 = -n$. $a_1 = \lambda, b_1 = m$ and $a_2 = \lambda, b_2 = -m$. Area $= \left| \frac{(0 - n)(0 - (-n))}{\lambda(-m) - \lambda(m)} \right| = \left| \frac{-n^2}{-2\lambda m} \right| = \frac{n^2}{2|\lambda m|}$ sq units.

The area (in sq units) of the quadrilateral formed by two pairs of lines $\lambda^2 x^2 - m^2 y^2 - n(\lambda x + m y) = 0$ and $\lambda^2 x^2 - m^2 y^2 + n(\lambda x + m y) = 0$ is:

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