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

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

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


Making 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 \theta be the required angle between these pair of lines, we have
Compare with general equation of 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}

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