# 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.

Prem Prakash

January 5@ 9:44 amtan∅ should be equal to ±2{√(h²-ab)}/(a+b).

As ± notation is missing out in the expression.

Kindly initiate the changes…