Question

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

1. $\dfrac{{{n^2}}}{{2|lm|}}$
2. $\dfrac{{{n^2}}}{{|lm|}}$
3. $\dfrac{n}{{2|lm|}}$
4. $\dfrac{{{n^2}}}{{4|lm|}}$

Answer 1

Hint : First, split each of the pairs of straight lines into two lines and analyse them. After analysing, we will conclude that they form a parallelogram with two sets of parallel lines. The Formula required for this question is that of finding the area of the parallelogram,$A = \dfrac{{|({c_1} - {d_1})({c_2} - {d_2})|}}{{|({a_1}{b_2} - {a_2}{b_1})|}}$. We must remember the general form of a linear equation in two variables as $ax + by + c = 0$.

Complete step-by-step answer :
Let’s first split the pair of straight lines into two by using the concept of factorization,$\Rightarrow {l^2}{x^2} - {m^2}{y^2} - n(lx + my) = 0$
Using the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$,
$\Rightarrow \left( {lx - my} \right)\left( {lx + my} \right) - n(lx + my) = 0$
Factoring common factors,
$\Rightarrow (lx + my)(lx - my - n) = 0$,
So, the two lines are
$\Rightarrow (lx + my) = 0$ and $\Rightarrow (lx - my - n) = 0$,
Now, repeating the same process for the other pair of straight line,
$\Rightarrow {l^2}{x^2} - {m^2}{y^2} + n(lx - my) = 0$
$\Rightarrow (lx - my)(lx + my + n) = 0$
So, the two lines are $\Rightarrow (lx - my) = 0$ and $\Rightarrow (lx + my + n) = 0$
Now, let’s analyse the four equations we get, a parallelogram is obtained with the 2 sides $(lx - my - n)$ and $(lx - my)$ parallel to each other and, $(lx + my)$ and $(lx + my + n)$ parallel to each other.
Now,
The area of the figure is given by $A = \dfrac{{|({c_1} - {c_2})({d_1} - {d_2})|}}{{|({m_1} - {m_2})|}}$, where ${c_1}$, ${c_2}$ are y intercepts of two parallel lines, ${d_1}$, ${d_2}$ are y intercepts of other two parallel lines, and ${m_1},{m_2}$ gives the slopes of parallel lines.
Now, we know that slope of parallel lines $(lx - my - n)$ and $(lx - my)$ is $\left( {\dfrac{l}{m}} \right)$ and y intercepts of both lines are $\left( {\dfrac{{ - n}}{m}} \right)$ and zero respectively.
Similarly, we know that slope of parallel lines $(lx + my)$ and $(lx + my + n)$ is $\left( { - \dfrac{l}{m}} \right)$ and y intercepts of both lines are zero and $\left( {\dfrac{{ - n}}{m}} \right)$ respectively.
Here, the parameters are given by ${c_1} = \left( {\dfrac{{ - n}}{m}} \right)$, ${c_2} = 0$, ${d_1} = {\text{\;0}}$, ${d_2} = \left( {\dfrac{{ - n}}{m}} \right)$, ${m_1} = \left( {\dfrac{l}{m}} \right)$ and ${m_2} = \left( { - \dfrac{l}{m}} \right)$.
After substituting these values, we get, $A = \left| {\dfrac{{\left( {\dfrac{{ - n}}{m} - 0} \right)\left( {0 - \left( {\dfrac{{ - n}}{m}} \right)} \right)}}{{\dfrac{l}{m} - \left( { - \dfrac{l}{m}} \right)}}} \right|$,
After solving the numerators and denominators separately, we get,
$\Rightarrow A = \left| {\dfrac{{\left( {\dfrac{{ - n}}{m}} \right)\left( {\dfrac{n}{m}} \right)}}{{\dfrac{l}{m} + \dfrac{l}{m}}}} \right|$
Simplifying the3 expression, we get,
$\Rightarrow A = \left| {\dfrac{{m{n^2}}}{{2l{m^2}}}} \right|$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow A = \left| {\dfrac{{{n^2}}}{{2lm}}} \right|$
Now, we know that any square of any number is positive. Therefore, it can get out of modulus with a positive sign, and similar as two is a constant, it comes out of the modulus operator with a positive sign. So, we get,
$A = \dfrac{{{n^2}}}{{2|lm|}}$
So, the correct answer is “Option 1”.

Note : This question is based on the concepts of pairs of straight lines and quadrilaterals. Be careful while splitting the pair of straight lines into two lines. We must know the properties of modulus function in order to get to the final answer. It will have two values , one positive and another negative. Do not commit calculation mistakes, and be sure of the final answer.
Area is the quantity that expresses the extent of a two-dimensional region, shape, or planar lamina, in the plane. Surface area is its analog on the two-dimensional surface of a three-dimensional object