Question

What is the angle between the straight lines $\left( {{m^2} - mn} \right)y = \left( {mn + {n^2}} \right)x + {n^3}$ and $\left( {mn + {m^2}} \right)y = \left( {mn - {n^2}} \right)x + {m^3}$ where, m>n?

Answer 1

Hint – First, we will find the slopes of both the equations and then put it in the formula to find the angle between them. Then by putting the values of slopes into the formula and further solving it we will get the value of $\theta $ i.e. angle between these two lines.

Complete step-by-step answer:
From the given equations
$\left( {{m^2} - mn} \right)y = \left( {mn + {n^2}} \right)x + {n^3}$ and $\left( {mn + {m^2}} \right)y = \left( {mn - {n^2}} \right)x + {m^3}$
We can write these equations as in slope – intercept form i.e. y=mx+b, where m is the slope i.e. coefficient of x.
 $y = \dfrac{{\left( {mn + {n^2}} \right)x + {n^3}}}{{\left( {{m^2} - mn} \right)}} = \dfrac{{\left( {mn + {n^2}} \right)x}}{{\left( {{m^2} - mn} \right)}} + \dfrac{{{n^3}}}{{\left( {{m^2} - mn} \right)}}$ and,
$y = \dfrac{{\left( {mn - {n^2}} \right)x + {m^3}}}{{\left( {mn + {m^2}} \right)}} = \dfrac{{\left( {mn - {n^2}} \right)x}}{{\left( {mn + {m^2}} \right)}} + \dfrac{{{m^3}}}{{\left( {mn + {m^2}} \right)}}$
First we will find the slopes of these given lines by comparing them with the slope – intercept form.
So, the slopes are:
${m_1} = \dfrac{{mn + {n^2}}}{{{m^2} - mn}}$ and ${m_2} = \dfrac{{mn - {n^2}}}{{mn + {m^2}}}$
Let $\theta $ be the angle between straight lines with slopes ${m_1}$ and ${m_2}$ is given by following formula:
$\tan \theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_2}{m_1}}}} \right|$
$ \Rightarrow \tan \theta = \left| {\dfrac{{\dfrac{{mn - {n^2}}}{{mn + {m^2}}} - \dfrac{{mn + {n^2}}}{{{m^2} - mn}}}}{{1 + \left( {\dfrac{{mn - {n^2}}}{{mn + {m^2}}}} \right)\left( {\dfrac{{mn + {n^2}}}{{{m^2} - mn}}} \right)}}} \right|$
$
   \Rightarrow \tan \theta = \left| {\dfrac{{\dfrac{{\left( {mn - {n^2}} \right)\left( {{m^2} - mn} \right) - \left( {mn + {n^2}} \right)\left( {mn + {m^2}} \right)}}{{\left( {{m^2} - mn} \right)\left( {mn + {m^2}} \right)}}}}{{1 + \left( {\dfrac{{\left( {mn - {n^2}} \right)\left( {mn + {n^2}} \right)}}{{\left( {mn + {m^2}} \right)\left( {{m^2} - mn} \right)}}} \right)}}} \right| \\
   \Rightarrow \tan \theta = \left| {\dfrac{{\dfrac{{\left( {mn - {n^2}} \right)\left( {{m^2} - mn} \right) - \left( {mn + {n^2}} \right)\left( {mn + {m^2}} \right)}}{{\left( {{m^2} - mn} \right)\left( {mn + {m^2}} \right)}}}}{{\dfrac{{\left( {mn + {m^2}} \right)\left( {{m^2} - mn} \right) + \left( {mn - {n^2}} \right)\left( {mn + {n^2}} \right)}}{{\left( {mn + {m^2}} \right)\left( {{m^2} - mn} \right)}}}}} \right| \\
 $
Now, on simplification we get,
$ \Rightarrow \tan \theta = \left| {\dfrac{{{m^3}n - {m^2}{n^2} - {m^2}{n^2} + m{n^3} - {m^2}{n^2} - {m^3}n - m{n^3} - {m^2}{n^2}}}{{{m^3}n - {m^2}{n^2} + {m^4} - {m^3}n + {m^2}{n^2} + m{n^3} - m{n^3} - {n^4}}}} \right|$
By cancelling out the same terms we get,
$
   \Rightarrow \tan \theta = \dfrac{{4{m^2}{n^2}}}{{{m^4} - {n^4}}} \\
   \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{4{m^2}{n^2}}}{{{m^4} - {n^4}}}} \right) \\
 $
Hence, the angle between the line is ${\tan ^{ - 1}}\left( {\dfrac{{4{m^2}{n^2}}}{{{m^4} - {n^4}}}} \right)$

Note – As you can see while solving for $\tan \theta $ student must pay attention while simplifying it as it is the most error prone step as it contains a number of terms. You have to follow this particular method only as you have to use the concept and formulas of slope-intercept form and angle between two lines.