Question

Calculate the area of the parallelogram formed by the lines \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,\text{ }{{a}_{1}}x+{{b}_{1}}y+{{d}_{1}}=0,\text{ }{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{d}_{2}}=0\] .
\[\begin{align}
  & \left( \text{a} \right)\text{ }\dfrac{\left( {{d}_{1}}-{{c}_{1}} \right)\left( {{d}_{2}}-{{c}_{2}} \right)}{{{\left[ \left( {{a}_{1}}^{2}+{{b}_{1}}^{2} \right)\left( {{a}_{2}}^{2}+{{b}_{2}}^{2} \right) \right]}^{\dfrac{1}{2}}}} \\
 & \left( b \right)\text{ }\dfrac{\left( {{d}_{1}}-{{c}_{1}} \right)\left( {{d}_{2}}-{{c}_{2}} \right)}{{{a}_{1}}{{a}_{2}}-{{b}_{1}}{{b}_{2}}} \\
 & \left( c \right)\text{ }\dfrac{\left( {{d}_{1}}+{{c}_{1}} \right)\left( {{d}_{2}}+{{c}_{2}} \right)}{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}} \\
 & \left( d \right)\text{ }\dfrac{\left( {{d}_{1}}-{{c}_{1}} \right)\left( {{d}_{2}}-{{c}_{2}} \right)}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\
\end{align}\]

Answer 1

- Hint: The basic formula we use is the area of the parallelogram that is
Area = base × height.
We can find the height by calculating the distance between parallel lines. Then we will find the points of intersection between a pair of parallel lines and a non-parallel line. By that we will get two points. The distance between them is the length of the base. We will multiply these two and get the answer.

Complete step-by-step solution -

Let us begin with the solution.
First, let us calculate the distance between the parallel line pair \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and\[{{a}_{1}}x+{{b}_{1}}y+{{d}_{1}}=0\].
Distance between parallel lines is given by the formula,
 \[=\dfrac{|{{d}_{1}}-{{c}_{1}}|}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}}\]
Therefore height between two parallel lines will be,
 \[=\dfrac{|{{d}_{1}}-{{c}_{1}}|}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}}\]
Now we need to find the intersection points between the pair of lines \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{d}_{2}}=0\] and the line \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] .

Let us start with \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] and \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] .
\[\begin{align}
  & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0..........\times {{a}_{1}} \\
 & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0..........\times {{a}_{2}} \\
\end{align}\]
\[\begin{align}
  & {{a}_{1}}{{a}_{2}}x+{{a}_{1}}{{b}_{2}}y+{{a}_{1}}{{c}_{2}}=0 \\
 & {{a}_{2}}{{a}_{1}}x+{{a}_{2}}{{b}_{1}}y+{{a}_{2}}{{c}_{1}}=0 \\
\end{align}\]
Now subtracting and cancelling the like terms, we get,
\[\begin{align}
  & {{a}_{1}}{{b}_{2}}y-{{a}_{2}}{{b}_{1}}y+{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}=0 \\
 & y({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})={{a}_{2}}{{c}_{1}}-{{a}_{1}}{{c}_{2}} \\
 & y=\dfrac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\
\end{align}\]
Now,
\[\begin{align}
  & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0..........\times {{b}_{1}} \\
 & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0..........\times {{b}_{2}} \\
\end{align}\]
\[\begin{align}
  & {{b}_{1}}{{a}_{2}}x+{{b}_{1}}{{b}_{2}}y+{{b}_{1}}{{c}_{2}}=0 \\
 & {{b}_{2}}{{a}_{1}}x+{{b}_{2}}{{b}_{1}}y+{{b}_{2}}{{c}_{1}}=0 \\
\end{align}\]
Subtracting and cancelling the like terms, we get,
\[\begin{align}
  & {{a}_{2}}{{b}_{1}}x-{{b}_{2}}{{a}_{1}}x+{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}=0 \\
 & x({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})={{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}} \\
 & x=\dfrac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\
\end{align}\]
Therefore, intersection point of \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] and \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] is \[\left( \dfrac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}},\dfrac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)\].
Let us start with \[{{a}_{2}}x+{{b}_{2}}y+{{d}_{2}}=0\] and \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\].
\[\begin{align}
  & {{a}_{2}}x+{{b}_{2}}y+{{d}_{2}}=0..........\times {{a}_{1}} \\
 & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0..........\times {{a}_{2}} \\
\end{align}\]
\[\begin{align}
  & {{a}_{1}}{{a}_{2}}x+{{a}_{1}}{{b}_{2}}y+{{a}_{1}}{{d}_{2}}=0 \\
 & {{a}_{2}}{{a}_{1}}x+{{a}_{2}}{{b}_{1}}y+{{a}_{2}}{{c}_{1}}=0 \\
\end{align}\]
Now subtracting and cancelling the like terms, we get,
\[\begin{align}
  & {{a}_{1}}{{b}_{2}}y-{{a}_{2}}{{b}_{1}}y+{{a}_{1}}{{d}_{2}}-{{a}_{2}}{{c}_{1}}=0 \\
 & y({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})={{a}_{2}}{{c}_{1}}-{{a}_{1}}{{d}_{2}} \\
 & y=\dfrac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{d}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\
\end{align}\]
Now,
\[\begin{align}
  & {{a}_{2}}x+{{b}_{2}}y+{{d}_{2}}=0..........\times {{b}_{1}} \\
 & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0..........\times {{b}_{2}} \\
\end{align}\]
\[\begin{align}
  & {{b}_{1}}{{a}_{2}}x+{{b}_{1}}{{b}_{2}}y+{{b}_{1}}{{d}_{2}}=0 \\
 & {{b}_{2}}{{a}_{1}}x+{{b}_{2}}{{b}_{1}}y+{{b}_{2}}{{c}_{1}}=0 \\
\end{align}\]
Subtracting and cancelling the like terms, we get,
\[\begin{align}
  & {{a}_{2}}{{b}_{1}}x-{{b}_{2}}{{a}_{1}}x+{{b}_{1}}{{d}_{2}}-{{b}_{2}}{{c}_{1}}=0 \\
 & x({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})={{d}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}} \\
 & x=\dfrac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{d}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\
\end{align}\]
Therefore, intersection point of \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] and \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] is \[\left( \dfrac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{d}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}},\dfrac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{d}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)\].
Now the distance between these two points is the base length.
Thus,
Base \[=\sqrt{{{\left( \dfrac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}-\dfrac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{d}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)}^{2}}+{{\left( \dfrac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}-\dfrac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{d}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)}^{2}}}\]
Since the LCM is the same in all we take it out. Cancelling the like terms, we get,
Base \[=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\sqrt{{{\left( {{b}_{1}}{{c}_{2}}+{{b}_{1}}{{d}_{2}} \right)}^{2}}+{{\left( -{{a}_{1}}{{c}_{2}}+{{a}_{1}}{{d}_{2}} \right)}^{2}}}\]
Taking out \[{{\left( {{c}_{2}}-{{d}_{2}} \right)}^{2}}\] as common we get,
Base \[=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\sqrt{\left( {{\left( {{b}_{1}} \right)}^{2}}+{{\left( {{a}_{1}} \right)}^{2}} \right){{\left( {{c}_{2}}-{{d}_{2}} \right)}^{2}}}\]
Taking out \[{{\left( {{c}_{1}}-{{d}_{1}} \right)}^{2}}\] out of the root we get,
Base \[=\dfrac{\left( {{c}_{2}}-{{d}_{2}} \right)}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\sqrt{\left( {{\left( {{b}_{1}} \right)}^{2}}+{{\left( {{a}_{1}} \right)}^{2}} \right)}\]
Now multiplying base and height we get,
Area \[=\dfrac{|{{d}_{1}}-{{c}_{1}}|}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}}\times \dfrac{|{{c}_{2}}-{{d}_{2}}|}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\sqrt{\left( {{\left( {{b}_{1}} \right)}^{2}}+{{\left( {{a}_{1}} \right)}^{2}} \right)}\]
Cancelling the common terms, we get,
Area \[=\dfrac{\left( {{d}_{1}}-{{c}_{1}} \right)\left( {{d}_{2}}-{{c}_{2}} \right)}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\].
Thus option(d) is correct.

Note: You might get confused as to why we include modulus sign always while we take out something out of the root. That is because we do not want our area value to be negative. Also the modulus sign eats up the negative sign and we can adjust our signs accordingly. Thus we get our final answer.