The angle bisector of L 1 : a 1 x + b 1 y + c 1 = 0and L 2 : a 2 x + b 2 y + c 2 = 0 ( a i , b i , c i ) ∈ R can be found be solving the equation a 1 x + b 1 y + c 1 a 1 2 + b 1 2 = ± a 2 x + b 2 y + c 2 a 2 2 + b 2 2 But our teacher told us that the equation of the angle bisector pasing through the region containing the origin can be obtained by solving only the positive case of the equation given that ( c 1 , c 2 ) &gt; 0.How can we prove this?

Question

The angle bisector of      L    1    :      a    1    x  +      b    1    y  +      c    1    =  0and      L    2    :      a    2    x  +      b    2    y  +      c    2    =  0  (      a    i    ,      b    i    ,      c    i    )  ∈      R  can be found be solving the equation                     a        1            x      +              b        1            y      +              c        1                            a        1        2            +              b        1        2              =  ±                       a        2            x      +              b        2            y      +              c        2                            a        2        2            +              b        2        2            But our teacher told us that the equation of the angle bisector pasing through the region containing the origin can be obtained by solving only the positive case of the equation given that   (      c    1    ,      c    2    )  &amp;gt;  0.How can we prove this?

Geovanni Shelton · Accepted Answer

|                          a          1                x        +                  b          1                y        +                  c          1                          |                                      a          1          2                +                  b          1          2                      =                            |                          a          2                x        +                  b          2                y        +                  c          2                          |                                      a          2          2                +                  b          2          2                          ⇒                            a          1                x        +                  b          1                y        +                  c          1                                      a          1          2                +                  b          1          2                      =    ±                                 a          2                x        +                  b          2                y        +                  c          2                                      a          2          2                +                  b          2          2                    Put (0,0) into both sides of your equation:                    c        1                              a          1          2                +                  b          1          2                      =    ±                       c        2                              a          2          2                +                  b          2          2                    You can see that both sides should yield nonnegative values, so we only choose   + in   ±.

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