Trigonometric question..Please help

sallyk57

New member
Joined
Feb 8, 2006
Messages
15
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.
 
Hello, sallyk57!

Line L\displaystyle L has slope m\displaystyle m, L1\displaystyle L_1 has slope m1\displaystyle m_1, and L2\displaystyle L_2 has slope [/tex]m_2[/tex]
Line L\displaystyle L bisects θ\displaystyle \theta, the angle formed by the lines L1\displaystyle L_1, and L2\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 m1\displaystyle m_1 and m2\displaystyle m_2,
the angle formed by the lines is given by: \(\displaystyle \L\,\tan\alpha\;=\;\frac{m_2\,-\,m_1}{1\,+\,m_1m_2}\)
Code:
                  *L1
                *
              *               *L
            *           *
          *  θ/2  *
        *   *   θ/2
      *   *   *   *   *    *    *L2
Since line L\displaystyle L bisects angle θ\displaystyle \theta,
the angle between L1\displaystyle L_1 and L\displaystyle L equals the angle between L\displaystyle L and L2\displaystyle L_2.


From the theorem:

    tanθ2  =  mm11+m1m\displaystyle \;\;\tan\frac{\theta}{2}\;= \;\frac{m\,-\,m_1}{1\,+\,m_1m}

    tanθ2  =  m2m1+m2m\displaystyle \;\;\tan\frac{\theta}{2}\;=\;\frac{m_2\,-\,m}{1\,+\,m_2m}

. . . Q.E.D.
 
Top