Two lines: a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 are given. I know that the equation of its bisectors is 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 I intend to find which one is the obtuse angle bisector and which one is the acute angle bisector. I want to find a general formula Assuming c 1 , c 2 both are of same sign, I know if a 1 a 2 + b 1 b 2 &gt; 0 and if we take the positive sign we get the obtuse angle bisector and vice versa. But I want to prove it using general equation of line, I tried to find the angle between bisector and original line i.e. t a n θ = m 1 − m 2 1 + m 1 m 2 and then if it is greater than one it will be of obtuse angle but calculations are tough if we use general equation of line. May anyone give a simple proof of the following statement: "Assuming c 1 , c 2 both are of same sign IF a 1 a 2 + b 1 b 2 &gt; 0 then if we take positive sign we get the obtuse angle bisector".

Question

Two lines:       a    1    x  +      b    1    y  +      c    1    =  0 and       a    2    x  +      b    2    y  +      c    2    =  0 are given. I know that the equation of its bisectors is                               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 I intend to find which one is the obtuse angle bisector and which one is the acute angle bisector. I want to find a general formula Assuming       c    1    ,      c    2   both are of same sign, I know if       a    1        a    2    +      b    1        b    2    &amp;gt;  0 and if we take the positive sign we get the obtuse angle bisector and vice versa. But I want to prove it using general equation of line, I tried to find the angle between bisector and original line i.e.   t  a  n      θ    =                              m          1                −                  m          2                            1        +                  m          1                          m          2                     and then if it is greater than one it will be of obtuse angle but calculations are tough if we use general equation of line. May anyone give a simple proof of the following statement: &quot;Assuming       c    1    ,      c    2   both are of same sign IF       a    1        a    2    +      b    1        b    2    &amp;gt;  0 then if we take positive sign we get the obtuse angle bisector&quot;.

Dobermann82 · Accepted Answer

We have two lines :      L    1    :      a    1    x  +      b    1    y  +      c    1    =  0  ,        L    2    :      a    2    x  +      b    2    y  +      c    2    =  0and the angle bisectors :      L          ±        :                    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            If we let   θ be the (smaller) angle between       L    +   and       L    1  , then we have  cos  ⁡  θ  =            |              a        1                    (                              a            1                                              a              1              2                        +                          b              1              2                                      −                              a            2                                              a              2              2                        +                          b              2              2                                      )            +              b        1                    (                              b            1                                              a              1              2                        +                          b              1              2                                      −                              b            2                                              a              2              2                        +                          b              2              2                                      )            |                                a          1          2                +                  b          1          2                                                  (                                          a                1                                                              a                  1                  2                                +                                  b                  1                  2                                                      −                                          a                2                                                              a                  2                  2                                +                                  b                  2                  2                                                      )                    2                +                              (                                          b                1                                                              a                  1                  2                                +                                  b                  1                  2                                                      −                                          b                2                                                              a                  2                  2                                +                                  b                  2                  2                                                      )                    2                      =            |                        a          1          2                +                  b          1          2                    −                                    a            1                                a            2                    +                      b            1                                b            2                                                a            2            2                    +                      b            2            2                              |                                a          1          2                +                  b          1          2                            2        −        2                                            a              1                                      a              2                        +                          b              1                                      b              2                                            (                          a              1              2                        +                          b              1              2                        )            (                          a              2              2                        +                          b              2              2                        )                                ×            2              1                              a            1            2                    +                      b            1            2                                      2              1                              a            1            2                    +                      b            1            2                                =                    1        −                                            a              1                                      a              2                        +                          b              1                                      b              2                                            (                          a              1              2                        +                          b              1              2                        )            (                          a              2              2                        +                          b              2              2                        )                              2      Hence, we can see that                              a          1                          a          2                +                  b          1                          b          2                &amp;gt;        0                                    ⟺                cos        ⁡        θ        &amp;lt;        1                  /                          2                                                          ⟺                θ        &amp;gt;                  45          ∘                                                          ⟺                                                    L              +                                 is the obtuse angle bisector                    as desired.(Note that &quot;      c    1    ,      c    2   both are of same sign&quot; is irrelevant.)

Two lines: a 1 </msub> x + b 1 </msub> y + c

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