Angle Between the Lines Represented by the Homogeneous Second Degree Equation

In previous tutorials, we saw 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 the 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 as

\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 in the form

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

Considering the above equation (ii) as a quadratic equation in terms of y and using the quadratic formula to solve 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, then

\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 an angle between the pair of straight lines.

(i) If the lines are parallel, then \theta = 0. So putting this value in the above equation, we have

{h^2} - ab = 0

(ii) If the lines are perpendicular, then \theta = {90^ \circ }. So putting this value in the above equation, we have

a + b = 0

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