Suppose you have a pair of lines passing through origin, a x 2 + 2 h x y + b y 2 = 0, how would you find the equation of pair of angle bisectors for this pair of lines. I can do this for 2 separate lines, but I am not able to figure it out for a pair of lines. Can someone please help.

Question

Suppose you have a pair of lines passing through origin,   a      x    2    +  2  h  x  y  +  b      y    2    =  0, how would you find the equation of pair of angle bisectors for this pair of lines. I can do this for 2 separate lines, but I am not able to figure it out for a pair of lines. Can someone please help.

Taniya Wood · Accepted Answer

Let&#039;s consider an easier case, when the given lines have the form   y  =  m  x and   y  =  n  x (with   m  ≠  n, of course).The points on the angle bisectors satisfy the conditions that their distance from the two lines is the same. So you have, for such a point   (  x  ,  y  ),                    |            y      −      m      x              |                    1      +              m        2              =                    |            y      −      n      x              |                    1      +              n        2            By squaring, we get  (  y  −  m  x      )    2    (  1  +      n    2    )  =  (  y  −  n  x      )    2    (  1  +      m    2    )and expanding  (  y  −  m  x      )    2    −  (  y  −  n  x      )    2    =      m    2    (  y  −  n  x      )    2    −      n    2    (  y  −  m  x      )    2    .The left hand side becomes  (  y  −  m  x  +  y  −  n  x  )  (  y  −  m  x  −  y  +  n  x  )  =  −  (  m  −  n  )  (  2  y  −  (  m  +  n  )  x  )  xand the right hand side is  (  m  y  −  m  n  x  +  n  y  −  m  n  x  )  (  m  y  −  m  n  x  −  n  y  +  m  n  x  )  =  (  m  −  n  )  (  (  m  +  n  )  y  −  2  m  n  x  )  ySince   m  ≠  n, we can factor out   m  −  n on both sides, getting  −  2  x  y  +  (  m  +  n  )      x    2    =  (  m  +  n  )      y    2    −  2  m  n  x  ythat can be written  (  m  +  n  )      x    2    −  2  (  1  −  m  n  )  x  y  −  (  m  +  n  )      y    2    =  0If you consider the equation   b      y    2    +  2  h  x  y  +  a      x    2    =  0 in the unknown   y, you can see that   m and   n are the roots of it (when   b  ≠  0). Thus   m  +  n  =  −  2  h      /    b and   m  n  =  a      /    b, so you can rewrite the above equation in the form            −      2      h        b        x    2    −  2      (    1    −          a      b        )    x  y  −            −      2      h        b        y    2    =  0which is  h      x    2    −  (  a  −  b  )  x  y  −  h      y    2    =  0Note that it can&#039;t be both   h  =  0 and   a  =  b  ≠  0, because in this case the original equations would be       x    2    +      y    2    =  0 (the isotropic lines, if you consider the complex plane).

Suppose you have a pair of lines passing through origin, a x 2 </msup> +

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