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.