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.

Answer 1

The given lines are

3x − 4y + a = 0 ... (1)

3x − 4y + 3a = 0 ... (2)

4x − 3y − a = 0 ... (3)

4x − 3y − 2a = 0 ... (4)



$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{parallelogram}=\left|\frac{\left(c_1-d_1\right)\left(c_2-d_2\right)}{a_1{b}_2-a_2{b}_1}\right| \\ \Rightarrow \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{parallelogram}=\left|\frac{\left(a-3a\right)\left(2a-a\right)}{-9+16}\right|=\frac{2a^2}{7}\mathrm{square}\mathrm{units} \end{gathered}$$