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

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(A) The given equations can be factored as follows:First pair: $(\lambda x - my)(\lambda x + my) - n(\lambda x + my) = 0 \implies (\lambda x + my)(\la

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given equations can be factored as follows:First pair: $(\lambda x - my)(\lambda x + my) - n(\lambda x + my) = 0 \implies (\lambda x + my)(\lambda x - my - n) = 0$.This represents the lines $L_1: \lambda x + my = 0$ and $L_2: \lambda x - my = n$.Second pair: $(\lambda x - my)(\lambda x + my) + n(\lambda x + my) = 0 \implies (\lambda x + my)(\lambda x - my + n) = 0$.This represents the lines $L_3: \lambda x + my = -n$ and $L_4: \lambda x - my = 0$.The area of the parallelogram formed by lines $a_1 x + b_1 y + c_1 = 0, a_1 x + b_1 y + c_2 = 0, a_2 x + b_2 y + d_1 = 0, a_2 x + b_2 y + d_2 = 0$ is given by $\left| \frac{(c_1 - c_2)(d_1 - d_2)}{a_1 b_2 - a_2 b_1} \right|$.Here,the lines are $\lambda x + my + 0 = 0, \lambda x + my + n = 0$ and $\lambda x - my + 0 = 0, \lambda x - my - n = 0$.Area $= \left| \frac{(0 - n)(0 - (-n))}{\lambda(-m) - m(\lambda)} \right| = \left| \frac{-n^2}{-2\lambda m} \right| = \frac{n^2}{2|\lambda m|}$ sq units.

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

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