Question: Find the angle between the lines represented by the following equation: x² + 5xy + 6y² = 0.
Solution:
Given,
Single equation of two straight lines is
x²+5xy+6y²=0
or, x² + 2×\frac{5}{2} xy + 6y²=0 - (i)
Comparing equation (i) with ax² + 2hxy + by²= 0, we get,
h = \frac{5}{2}, a = 1, b = 6
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{5}{2})^2 - 1×6}}{1 +6} \right )
or, tan \theta = \left ( \pm \dfrac{ 2\sqrt{\frac{25}{4} - 6}}{7} \right )
or, tan \theta = \left ( \pm \dfrac{2 \sqrt{ \frac{25 -24}{4}}}{7} \right )
or, tan \theta = \left ( \pm \dfrac{2 \sqrt{ \frac{1}{4}}}{7} \right )
or, tan \theta = \left ( \pm \dfrac{2 × \frac{1}{2}}{7} \right )
or, tan \theta = \left ( \pm \dfrac{1}{7} \right )
Taking positive sign,
or, tan \theta = \frac{1}{7}
or, \theta = tan^{-} \frac{1}{7}
\therefore \theta = 8° [Use calculator]
Taking negative sign,
or, tan \theta = - \frac{1}{7}
or, \theta = tan^{-1} ( - \frac{1}{7} )
\therefore \theta = 172° [Use calculator]
Therefore, the required angle between two lines represented by the given single equation of two lines is 8° or 172°.
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