Question

Find the angle between the pairs of straight lines: $\left( {{m^2} - mn} \right)y = (mn + {n^2})x + {n^3}$ and $\left( {mn + {m^2}} \right)y = \left( {mn - {n^2}} \right)x + {m^3}$.

Answer 1

Hint: In this question first we will find the slope of the two lines given in the question by comparing the equation of lines with the general equation of line. Now, with the help of the formula of angle between two lines we can easily find the angle between the pairs of straight lines.

Complete step by step solution: Given pairs of straight lines are: $\left( {{m^2} - mn} \right)y = (mn + {n^2})x + {n^3}$ and $\left( {mn + {m^2}} \right)y = \left( {mn - {n^2}} \right)x + {m^3}$
Now, we can write the above equation as
$y = \dfrac{{(mn + {n^2})x}}{{\left( {{m^2} - mn} \right)}} + \dfrac{{{n^3}}}{{\left( {{m^2} - mn} \right)}}\,\,\_\_\_\left( 1 \right)$ and $y = \dfrac{{\left( {mn - {n^2}} \right)x}}{{\left( {mn + {m^2}} \right)}} + \dfrac{{{m^3}}}{{\left( {mn + {m^2}} \right)}}\,\,\_\_\_\left( 2 \right)$
Now, we will compare the equation $\left( 1 \right)$ and equation $\left( 2 \right)$ with the general equation of straight line $y = mx + c$, where $m$ is the slope of the line and $c$ is the intercept on the $y$ axis.
For the equation $\left( 1 \right)$ we got the value of slope ${m_1}$
${m_1} = \dfrac{{mn + {n^2}}}{{{m^2} - mn}}$
For the equation $\left( 2 \right)$ we got the value of slope ${m_2}$
${m_2} = \dfrac{{mn - {n^2}}}{{mn + {m^2}}}$
Now, with the help of the formula of angle between two straight lines. We can find the angle:
$\tan \theta = \pm \left( {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right)$ , where ${m_1}$ and ${m_2}$ are slopes
Now, put the values of slopes in the above equation:
$ \Rightarrow \tan \theta = \left( {\dfrac{{\dfrac{{mn + {n^2}}}{{{m^2} - mn}} - \dfrac{{mn - {n^2}}}{{mn + {m^2}}}}}{{1 + \left( {\dfrac{{mn + {n^2}}}{{{m^2} - mn}}} \right)\left( {\dfrac{{mn - {n^2}}}{{mn + {m^2}}}} \right)}}} \right)$
The above equation can be written as:
$ \Rightarrow \tan \theta = \left( {\dfrac{{\dfrac{{\left( {mn + {n^2}} \right)\left( {mn + {m^2}} \right) - \left( {mn - {n^2}} \right)\left( {{m^2} - mn} \right)}}{{\left( {{m^2} - mn} \right)\left( {mn + {m^2}} \right)}}}}{{\dfrac{{\left( {{m^2} - mn} \right)\left( {mn + {m^2}} \right) + \left( {mn + {n^2}} \right)\left( {mn - {n^2}} \right)}}{{\left( {{m^2} - mn} \right)\left( {mn + {m^2}} \right)}}}}} \right)$
Now, we can simplify the above equation as:
$ \Rightarrow \tan \theta = \dfrac{{\left( {mn + {n^2}} \right)\left( {mn + {m^2}} \right) - \left( {mn - {n^2}} \right)\left( {{m^2} - mn} \right)}}{{\left( {{m^2} - mn} \right)\left( {mn + {m^2}} \right) + \left( {mn + {n^2}} \right)\left( {mn - {n^2}} \right)}}$
Open the brackets in numerator and denominator and simplify the equation.
$ \Rightarrow \tan \theta = \dfrac{{{{\left( {mn} \right)}^2} + {m^3}n + m{n^3} + {n^2}{m^2} - \left( {{m^3}n - {{\left( {mn} \right)}^2} - {m^2}{n^2} + m{n^3}} \right)}}{{{m^3}n + {m^4} - {{\left( {mn} \right)}^2} - {m^3}n + {{\left( {mn} \right)}^2} - {n^4}}}$
Again, simplify the above equation:
$ \Rightarrow \tan \theta = \dfrac{{{m^2}{n^2} + {m^3}n + m{n^3} + {n^2}{m^2} - {m^3}n + {m^2}{n^2} + {m^2}{n^2} - m{n^3}}}{{{m^4} - {n^4}}}$
$\Rightarrow \tan \theta = \dfrac{{4{m^2}{n^2}}}{{{m^4} - {n^2}}} $
Now, the angle between the pairs of straight lines can be written as:
$ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{4{m^2}{n^2}}}{{{m^4} - {n^2}}}} \right)$
Hence, we found the angle between two straight lines $\theta = {\tan ^{ - 1}}\left( {\dfrac{{4{m^2}{n^2}}}{{{m^4} - {n^2}}}} \right)$

Note: In this question the important thing is how we can convert the equation of pairs of straight lines into the form of general equation of straight line. The other important thing is finding the slope by comparing the given equation with the general equation of the straight line.