The given lines are
![\[\begin{array}{*{35}{l}} 3x~-~4y\text{ }+\text{ }a\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 1 \right) \\ 3x~-~4y\text{ }+\text{ }3a\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 2 \right) \\ 4x~-~3y~-~a\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 3 \right) \\ 4x~-~3y~-~2a\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 4 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-202c55d5d792ff3a761a0393752ad0df_l3.png "Rendered by QuickLaTeX.com")
Let us prove, 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 2a2/7 sq. units.
From above solution, we know that
Hence proved.
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 2a2/7 sq. units.
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 2a2/7 sq. units.