First slide

Equation of line in Straight lines

Question

If the area of the parallelogram formed b:1 the lines $2 x - 3 y + a = 0, 3 x - 2 y - a = 0, 2 x - 3 y + 3 a = 0$ and $3 x - 2 y - 2 a = 0$ is 10 square units, then $| a | =$

Moderate

Solution

We know that the area of the parallelogram iormed by the lines $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$
$a_{1} x + b_{1} y + d_{1} = 0 and a_{2} x + b_{2} y + d_{2} = 0$ is
$\begin{array}{l} |\frac{(c_{1} - d_{1}) (c_{2} - d_{2})}{|\begin{matrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{matrix}|}| \\ |\frac{(3 a - a) \times (- 2 a + a)}{|\begin{array}{ll} 2 & - 3 \\ 3 & - 2 \end{array}|}| = 10 \Rightarrow \frac{2 a^{2}}{5} = 10 \Rightarrow a = \pm 5 \end{array}$
Hence $| a | =$ 5

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