# 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.