Question: Show that the equation 10x² -11xy -6y² -12x +18y = 0 represents two straight lines. Find the angle between them.

Solution:
Given,

Single equation of two straight lines is
10x² -11xy -6y² -12x +18y = 0
or, 10x² - (15-4)xy -6y² -6(2x -3y) = 0
or, 10x² -15xy +4xy -6y² -6(2x -3y) = 0
or, 5x(2x -3y) +2y(2x -3y) -6(2x -3y) = 0
or, (5x +2y -6)(2x -3y) = 0

So, equation of line 1 is (5x +2y -6 = 0)
And, equation of line 2 is (2x -3y = 0).

Also,
Slope of line 1 (5x +2y -6 = 0) (m_1) = - \frac{coefficient \; of \; x}{coefficient \; of y}
= - \frac{5}{2}

Slope of line 2 (2x -3y = 0) (m_2) = - \frac{coefficient \; of \; x}{coefficient \; of y}
= - \frac{2}{-3}
= \frac{2}{3}

Now,
Let the angle between line 1 and line 2 be \theta, we have,
or, tan \theta = \pm \dfrac{m_1 - m_2}{1 + m_1m_2}

or, tan \theta = \pm \dfrac{-\frac{5}{2} - \frac{2}{3} } { 1 + ( - \frac{5}{2}) × \frac{2}{3} }

or, tan \theta = \pm \dfrac{- ( \frac{15 +4}{6} )}{1 - \frac{5}{3}}

or, tan \theta = \pm \dfrac{- \frac{19}{6}}{\frac{3-5}{3}

or, tan \theta = \pm \dfrac{- \frac{19}{6}}{\frac{-2}{3}}

or, tan \theta = \pm \left ( \frac{-19}{6} ×  \frac{3}{-2} \right )

or, tan \theta = \pm \frac{-57}{-12}

or, tan \theta = \pm \frac{57}{12}

Taking positive sign,
or, tan \theta = \frac{57}{12}
or, \theta = tan^{-1} (4.75)
\therefore \theta = 78.11°

Taking negative sign,
or, tan \theta = - \frac{57}{12}
or, tan \theta = - 4.75
or, \theta = tan^{-1} (-4.75)
\therefore \theta = 101.88°

Hence, the required angle between the two lines is either 78.11° or 101.88°.

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