# Example of Finding the Angle Between the Lines Represented by the Second Degree Homogeneous Equation

Find the lines represented by the second degree homogeneous equation $$3{x^2} + 7xy + 2{y^2} = 0$$. Also find the measure of the angle between them.

We have the second degree homogeneous equation:

\[3{x^2} + 7xy + 2{y^2} = 0\]

Calculating the factors of the given equation:

\[\begin{gathered} 3{x^2} + 6xy + xy + 2{y^2} = 0 \\ \Rightarrow 3x\left( {x + 2y} \right) + y\left( {x + 2y} \right) = 0 \\ \Rightarrow \left( {3x + y} \right)\left( {x + 2y} \right) = 0 \\ \Rightarrow 3x + y = 0,\,\,\,\, \Rightarrow x + 2y = 0 \\ \end{gathered} \]

This represents the required equations of straight lines passing through the origin.

Now let $$\theta $$ be the required angle between this pair of lines, and we have

Compare with the general equation of a homogeneous equation $$a{x^2} + 2hxy + b{y^2} = 0$$.

Here $$a = 3,\,\,b = 2,\,\,h = \frac{7}{2}$$. If $$\theta $$ is the angle between the pair of lines, then

\[\begin{gathered} \tan \theta = \frac{{2\sqrt {{h^2} – ab} }}{{a + b}} = \frac{{2\sqrt {\frac{{49}}{4} – 6} }}{{3 + 2}} \\ \Rightarrow \tan \theta = \frac{{\sqrt {25} }}{5} = \frac{5}{5} = 1 \\ \Rightarrow \theta = {45^ \circ } \\ \end{gathered} \]