Question: Find the angle between the lines represented by the following equation: 5x² + 7xy + 2y² = 0.

Solution:
Given,

Single equation of two straight lines is 
$5x²+7xy+2y²=0$
$or, 5x² + 2×\frac{7}{2} xy + 2y²=0$ - (i)

Comparing equation (i) with ax² + 2hxy + by²= 0, we get,
$h = \frac{7}{2}, a = 5, b = 2$

Angle represented by the following single equation is given by:

$tan \theta = \left ( \pm \dfrac{2 \sqrt{h^2 - ab}}{a +b} \right )$

$or, tan \theta = \left ( \pm \dfrac{2 \sqrt{ (\frac{7}{2})^2 - 5×2}}{5 +2} \right )$

$or, tan \theta = \left ( \pm \dfrac{ 2\sqrt{\frac{49}{4} - 10}}{7} \right )$

$or, tan \theta = \left ( \pm \dfrac{2 \sqrt{ \frac{49-40}{4}}}{7} \right )$

$or, tan \theta = \left ( \pm \dfrac{2 \sqrt{ \frac{9}{4}}}{7} \right )$

$or, tan \theta = \left ( \pm \dfrac{2 × \frac{3}{2}}{7} \right )$

$or, tan \theta = \left ( \pm \dfrac{3}{7} \right )$

Taking positive sign,
$or, tan \theta = \frac{3}{7}$
$or, \theta = tan^{-} \frac{3}{7}$
$\therefore \theta = 23.2°$ [Use calculator]

Taking negative sign,
$or, tan \theta = - \frac{3}{7}$
$or, \theta = tan^{-1} ( - \frac{3}{7} )$
$\therefore \theta = 156.8°$ [Use calculator]

Therefore, the required angle between two lines represented by the given single equation of two lines is 23.2° or 156.8°.