Angle between the Lines Represented by Homogeneous Second Degree Equation

We studied in previous tutorials that the equation of the form a{x^2} + 2hxy + b{y^2} = 0 is called the second degree homogeneous equation. And we know that second degree homogeneous equation represents the pair of straight lines passing through the origin.

Now the angle \theta between the lines represented by the homogeneous second degree equation as a{x^2} + 2hxy + b{y^2} = 0 is given by

\tan  \theta  = \frac{{2\sqrt {{h^2} - ab}  }}{{a + b}}


The second degree homogeneous equation is given as

a{x^2}  + 2hxy + b{y^2} = 0\,\,\,\,{\text{ -  -  - }}\left( {\text{i}} \right)


This equation (i) can be rewritten of the form

b{y^2}  + 2hxy + a{x^2} = 0\,\,\,\,{\text{ -   -  - }}\left( {{\text{ii}}}  \right)


Considering the above equation (ii) as quadratic equation in terms of y and using quadratic formula for solving this equation, we have

\begin{gathered} y = \frac{{ - 2hx \pm \sqrt {{{\left( {2hx}  \right)}^2} - 4\left( b \right)\left( {a{x^2}} \right)} }}{{2\left( b \right)}} \\ \Rightarrow y = \frac{{2x\left( { - h \pm  \sqrt {{h^2} - ab} } \right)}}{{2b}} \\ \Rightarrow y = \left( {\frac{{ - h \pm  \sqrt {{h^2} - ab} }}{b}} \right)x\,\,\,{\text{ -  -  -  }}\left( {{\text{iii}}} \right) \\ \end{gathered}


Let {m_1} = \frac{{ - h  + \sqrt {{h^2} - ab} }}{b} and {m_2}  = \frac{{ - h - \sqrt {{h^2} - ab} }}{b}
Making these substitutions, equations (iii) are y = {m_1}x and y = {m_2}x which is the pair of lines represented by the given homogeneous second degree equation.
Now

\begin{gathered} {m_1} + {m_2} = \frac{{ - h + \sqrt {{h^2} -  ab} }}{b} + \frac{{ - h - \sqrt {{h^2} - ab} }}{b} \\ \Rightarrow {m_1} + {m_2} = \frac{{ - h +  \sqrt {{h^2} - ab}  - h - \sqrt {{h^2} -  ab} }}{b} = \frac{{ - 2h}}{b} \\ \end{gathered}


Also

\begin{gathered} {m_1}{m_2} = \left( {\frac{{ - h + \sqrt  {{h^2} - ab} }}{b}} \right)\left( {\frac{{ - h - \sqrt {{h^2} - ab} }}{b}}  \right) \\ {m_1}{m_2} = \frac{{{{\left( { - h}  \right)}^2} - {{\left( {\sqrt {{h^2} - ab} } \right)}^2}}}{{{b^2}}} =  \frac{a}{b} \\ \end{gathered}


Since \theta is the angle between the lines, so

\tan  \theta  = \frac{{{m_2} - {m_1}}}{{1 +  {m_1}{m_2}}} = \frac{{\sqrt {{{\left( {{m_1} - {m_2}} \right)}^2}} }}{{1 +  {m_1}{m_2}}} = \frac{{\sqrt {{{\left( {{m_1} + {m_2}} \right)}^2} - 4{m_1}{m_2}}  }}{{1 + {m_1}{m_2}}}


Using the values of {m_1} and {m_2} in the above equation, we have

\begin{gathered} \tan \theta   = \frac{{\sqrt {{{\left( { - \frac{{2h}}{b}} \right)}^2} - 4\left(  {\frac{a}{b}} \right)} }}{{1 + \left( {\frac{a}{b}} \right)}} \\ \Rightarrow \tan \theta  = \frac{{2\sqrt {{h^2} - ab} }}{{a + b}} \\ \end{gathered}


This completes the proof for angle between the pair of straight lines.
(i) If the line are parallel, then \theta   = 0, so putting this values in the above equation, we have

{h^2}  - ab = 0


(ii) If the line are perpendicular, then \theta  = {90^ \circ }, so putting this values in the above equation, we have

a +  b = 0


This is the condition for two lines to perpendicular to each other.

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