Trigonometric question..Please help

[sallyk57]

Line l has slope m, l1 has slope m1, and l2 has slope m2. Line l bisects theta, the angle formed by the lines l1, and l2. Show that:

(m1 - m )/ (1+m1m) = (m - m2)/ (1+m2m)

Also, If l1 and l2 have equations y=2x and y=x, find an equation of l.


I know that tan theta = (m1 -m2) / (1+ m1m2). But, I don't know how to start.

 

[soroban]

Hello, sallyk57!

Line \(\displaystyle L\) has slope \(\displaystyle m\), \(\displaystyle L_1\) has slope \(\displaystyle m_1\), and \(\displaystyle L_2\) has slope [/tex]m_2[/tex]
Line \(\displaystyle L\) bisects \(\displaystyle \theta\), the angle formed by the lines \(\displaystyle L_1\), and \(\displaystyle L_2\)

Show that: \(\displaystyle \L\,\frac{m_1\,-\,m}{1\,+\,m_1m} \;= \;\frac{m\,-\,m_2}{1\,+\,m_2m}\)

Theorem: given two lines with slopes \(\displaystyle m_1\) and \(\displaystyle m_2\),
the angle formed by the lines is given by: \(\displaystyle \L\,\tan\alpha\;=\;\frac{m_2\,-\,m_1}{1\,+\,m_1m_2}\)

                  *L1
                *
              *               *L
            *           *
          *  θ/2  *
        *   *   θ/2
      *   *   *   *   *    *    *L2

Since line \(\displaystyle L\) bisects angle \(\displaystyle \theta\),
the angle between \(\displaystyle L_1\) and \(\displaystyle L\) equals the angle between \(\displaystyle L\) and \(\displaystyle L_2\).


From the theorem:

\(\displaystyle \;\;\tan\frac{\theta}{2}\;= \;\frac{m\,-\,m_1}{1\,+\,m_1m}\)

\(\displaystyle \;\;\tan\frac{\theta}{2}\;=\;\frac{m_2\,-\,m}{1\,+\,m_2m}\)

. . . Q.E.D.