# The Second Degree Homogeneous Equation Represents a Pair of Lines

As we know that the equation of the form $$a{x^2} + 2hxy + b{y^2} = 0$$ is called the second degree homogeneous equation, the second degree homogeneous equation represents the pair of straight lines passing through the origin.

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

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

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 are obviously equations of lines passing through the origin because there are no y-intercepts in these equations (iii).

This shows that the second degree homogeneous equation represents the pair of straight lines passing through the origin.