1.0 Definition

An equation of the form $${a_0}{y^n} + {a_1}{y^{n - 1}}x + {a_2}{y^{n - 2}}{x^2} + .... + {a_n}{x^n} = 0...(1)$$ in which the sum of the powers of $x$ and $y$ in every term is the same (here $n$), is called a homogeneous equation (of degree $n$).

Divide each term in equation $(1)$ by ${x^n}$, we get $${a_0}{(\frac{y}{x})^n} + {a_1}{(\frac{y}{x})^{n - 1}} + {a_2}{(\frac{y}{x})^{n - 2}} + .... + {a_n} = 0$$ which is an equation of $n$th degree in $\frac{y}{x}$ and the roots of this equation be ${m_1},{m_2},{m_3},....,{m_n}$. Therefore, the above equation is similar to $${a_0}(\frac{y}{x} - {m_1})(\frac{y}{x} - {m_2})(\frac{y}{x} - {m_3})....(\frac{y}{x} - {m_n}) = 0$$ or, $${a_0}(y - {m_1}x)(y - {m_2}x)(y - {m_3}x)....(y - {m_n}x) = 0$$

Hence, equation $(1)$ represents $n$ straight lines which pass through origin.

Similarly, the equation $$a{x^2} + 2hxy + b{y^2} = 0...(2) $$ represents a homogeneous equation of second degree i.e., represents two straight lines passing through origin. Divide equation $(2)$ by ${x^2}$, we get $$\begin{equation} \begin{aligned} a + 2h(\frac{y}{x}) + b{(\frac{y}{x})^2} = 0 \\ b{(\frac{y}{x})^2} + 2h(\frac{y}{x}) + a = 0 \\\end{aligned} \end{equation} $$

Putting $\frac{y}{x} = m$, we get $$b{m^2} + 2hm + a = 0$$

which is a quadratic equation in $m$. Let ${m_1}$ and ${m_2}$ be two roots of the equation, then

$$\begin{equation} \begin{aligned} {m_1} + {m_2} = - \frac{{2h}}{b} = - \frac{{{\text{coefficient of }}xy}}{{{\text{coefficient of }}{y^2}}}...(3) \\ {\text{and}} \\ {m_1} \times {m_2} = \frac{a}{b} = \frac{{{\text{coefficient of }}{x^2}}}{{{\text{coefficient of }}{y^2}}}...(4) \\\end{aligned} \end{equation} $$

From equation $(3)$ and $(4)$,

$$\left| {{m_1} - {m_2}} \right| = \sqrt {\{ {{({m_1} + {m_2})}^2} - 4{m_1}{m_2}\} } = \frac{2}{b}\sqrt {({h^2} - ab)}...(5) $$

Thus, $y = {m_1}x$ and $y = {m_2}x$ are two straight lines represented by equation $(2)$.

Now, consider the equation $$b{m^2} + 2hm + a = 0$$

We can write $$\begin{equation} \begin{aligned} m = \frac{{ - 2h \pm 2\sqrt {({h^2} - ab)} }}{{2b}} = \frac{{ - h \pm \sqrt {({h^2} - ab)} }}{b} = \frac{y}{x} \\ \therefore by = \{ - h + \sqrt {({h^2} - ab)} \} x{\text{ and }}by = \{ - h - \sqrt {({h^2} - ab)} \} x \\\end{aligned} \end{equation} $$ are the equations of two lines represented by equation $(2)$.