Question

Find what straight lines are represented by the following equation and determine the angles between them.
$33x^2-71xy-14y^2=0$

Answer 1

${33x}^2-71xy-14y^2=0$

Factorising the equation

${33x}^2-77xy+6xy-14y^2=011x(3x-7y)+2y(3x-7y)=0(11x+2y)(3x-7y)=011x+2y=0,3x-7y=0\Rightarrow y=\frac{-11}{2}x,y=\frac{3}{7}x$

Slope of first line $m_1=\frac{-11}{2}$

Slope of second line $m_2=\frac{3}{7}$

Angle between two straight lines

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\frac{-11}{2}-\frac{3}{7}}{1+\left(\frac{-11}{2}\right).\frac{3}{7}}\right|\tan \theta =\frac{\frac{83}{14}}{\frac{19}{14}}=\frac{83}{19}\theta ={\tan }^{-1}\left(\frac{83}{19}\right)$