Question

The area (in square units) of the quadrilateral formed by the two pairs of lines$l^2{x}^2-m^2{y}^2-n(lx+my)=0$ and$l^2{x}^2-m^2{y}^2+n(lx-my)=0$

• A. $\frac{n^2}{2\left|lm\right|}$
• B. $\frac{n^2}{\left|lm\right|}$
• C. $\frac{n}{2\left|lm\right|}$
• D. $\frac{n^2}{4\left|lm\right|}$

Answer 1

The correct option is A

$\frac{n^2}{2\left|lm\right|}$

Explanation for correct option:

Given two pairs of lines are,

$$\begin{gathered} l^2{x}^2-m^2{y}^2-n(lx+my)=0 \\ \Rightarrow \left(lx-my\right)\left(lx+my\right)-n\left(lx+my\right)=0\left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ \Rightarrow \left(lx+my\right)\left(lx-my-n\right)=0 \end{gathered}$$

Hence the two lines are

$$\begin{gathered} lx+my=0\left(1\right) \\ lx-my-n=0\left(2\right) \end{gathered}$$

Also,

$$\begin{gathered} l^2{x}^2-m^2{y}^2+n(lx-my)=0 \\ \Rightarrow \left(lx-my\right)\left(lx+my\right)+n\left(lx-my\right)=0 \\ \Rightarrow \left(lx-my\right)\left(lx+my+n\right)=0 \end{gathered}$$

Hence the two lines are

$$\begin{gathered} lx-my=0\left(3\right) \\ lx+my+n=0\left(4\right) \end{gathered}$$

Now the area of a quadrilateral formed by four lines is given by,

$$\begin{gathered} \mathrm{Area}=\left|\frac{\left(c_1-d_1\right)\left(c_2-d_2\right)}{a_1{b}_2-a_2{b}_1}\right| \\ =\left|\frac{\left(0-n\right)\left(0+n\right)}{-lm-lm}\right| \\ =\frac{n^2}{2\left|lm\right|} \end{gathered}$$

Therefore the area of a quadrilateral formed is equal to $\frac{n^2}{2\left|lm\right|}$.

Hence, the correct answer is Option (A).