Second Degree Homogeneous Equation Represents 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 of the form

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


Considering the above equation (ii) as quadratic equation in terms of y and using quadratic formula for solving 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 origin because there is no y-intercepts in these equation (iii).
This shows that the second degree homogeneous equation represents the pair of straight lines passing through the origin.

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