Question

Prove that the area of the parallelogram formed by the lines a₁x + b₁y + c₁ = 0, a₁x + b₁y+ d₁ = 0, a₂x + b₂y + c₂ = 0, a₂x + b₂y + d₂ = 0 is  \[\left| \frac{\left( d_1 - c_1 \right)\left( d_2 - c_2 \right)}{a_1 b_2 - a_2 b_1} \right|\] sq. units.
Deduce the condition for these lines to form a rhombus.

 

Answer 1

The given lines are
a₁x + b₁y + c₁ = 0      ... (1)
a₁x + b₁y + d₁ = 0      ... (2)
a₂x + b₂y + c₂ = 0      ... (3)
a₂x + b₂y + d₂ = 0      ... (4)
The area of the parallelogram formed by the lines a₁x + b₁y + c₁ = 0, a₁x + b₁y + d₁ = 0, a₂x + b₂y + c₂ = 0 and a₂x + b₂y + d₂ = 0 is given below:

\[Area = \left| \frac{\left( c_1 - d_1 \right)\left( c_2 - d_2 \right)}{\begin{vmatrix}a_1 & a_2 \\ b_1 & b_2\end{vmatrix}} \right|\]

\[\because \begin{vmatrix}a_1 & a_2 \\ b_1 & b_2\end{vmatrix} = a_1 b_2 - a_2 b_1\]

\[\therefore Area = \left| \frac{\left( c_1 - d_1 \right)\left( c_2 - d_2 \right)}{a_1 b_2 - a_2 b_1} \right| = \left| \frac{\left( d_1 - c_1 \right)\left( d_2 - c_2 \right)}{a_1 b_2 - a_2 b_1} \right|\]

If the given parallelogram is a rhombus, then the distance between the pair of parallel lines are equal.

\[\therefore \left| \frac{c_1 - d_1}{\sqrt{{a_1}^2 + {b_1}^2}} \right| = \left| \frac{c_2 - d_2}{\sqrt{{a_2}^2 + {b_2}^2}} \right|\]