Question

# Prove that the area of the parallelogram formed by the lines 3x − 4y + a = 0, 3x − 4y + 3a = 0, 4x − 3y − a = 0 and 4x − 3y − 2a = 0 is $\frac{2}{7}{a}^{2}$ sq. units.

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Solution

## The given lines are 3x − 4y + a = 0 ... (1) 3x − 4y + 3a = 0 ... (2) 4x − 3y − a = 0 ... (3) 4x − 3y − 2a = 0 ... (4) $\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{parallelogram}=\left|\frac{\left({c}_{1}-{d}_{1}\right)\left({c}_{2}-{d}_{2}\right)}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}\right|\phantom{\rule{0ex}{0ex}}⇒\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{parallelogram}=\left|\frac{\left(a-3a\right)\left(2a-a\right)}{-9+16}\right|=\frac{2{a}^{2}}{7}\mathrm{square}\mathrm{units}$

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