Find the tangent of the acute angle between the lines \(2 x+y=3\) and \(3 x-2 y=5\)
Explanation
Key: the tangent of the acute angle between two lines is given as
\(\tan \alpha=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|\)
Notice the use of the absolute value to ensure that \(\tan \theta\) is equal to a positive number.or the line \(2 x+y=3\)
\(y=-2 x+3\)
\(m_1=-2\) and for the line \(3 x-2 y=5\)
\(-2 y=-3 x+5\)
\(y=\frac{3}{2} x-\frac{5}{2}\)
\(m_2=+\frac{3}{2}\)
so, \(\tan \alpha=\left|\frac{\frac{3}{2}-(-2)}{1+(-2)\left(\frac{3}{2}\right)}\right|\)
\(\tan \alpha=\left|\frac{2}{2}+2\right|=\frac{7}{1-3} \mid \frac{2}{-2}\)
\(\tan a=\left|\frac{-7}{4}\right|=\frac{7}{4}\)