Equations of the bisectors of the angle between lines

Equation of the Bisectors:

The bisectors of the angle between two straight lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are the locus of a point which is equidistant from the two lines.

So let P (h, k) be a point equidistant from the two lines. Then, PL = PM.

⇒ | [(a₁h + b₁k + c₁)/ √ (a₁² + b₁²)]| = | [(a₂h + b₂k + c₂)/ √ (a₂² + b₂²)] |

∴ Locus of (h, k) i.e., the equation of the bisectors are | (a₁h + b₁k + c₁)|/ √ (a₁² + b₁²) = | (a₂h + b₂k + c₂)|/ √ (a₂² + b₂²)

⇒ a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 ⇒ (a₁x + b₁y + c₁)/ √ (a₁² + b₁²) = ± (a₂x + b₂y + c₂)/ √ (a₂² + b₂²)

These are the required equations of bisectors.

Example:

Find the equations of the bisectors of the angles between the straight lines 3x – 4y + 7 = 0 and 12x – 5y – 8 = 0.

Solution:

The equation of the bisectors of the angles between 3x – 4y + 7 = 0 and 12x – 5y – 8 = 0 are (3x – 4y + 7)/ √ (3² + (- 4)²) = ± (12x – 5y – 8)/ √ (12² + (- 5)²)

Or, (3x – 4y + 7)/ 5 = ± (12x – 5y – 8)/ 13

Or, 39x – 52y + 91 = ± (60x – 25y - 40)

Taking the positive sign, we get

21x + 27y – 131 = 0 as one bisector.

Taking the negative sign, we get

99x – 77y + 51 = 0 as the other bisector.

Bisector of the angle containing the origin:

Let the equation of two lines be a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 where c₁ and c₂ are positive. Then the equation of bisector of the angle containing origin is (a₁x + b₁y + c₁)/ √ (a₁² + b₁²) = (a₂x + b₂y + c₂)/ √ (a₂² + b₂²)

Algorithm:

Step 1: Obtain the two lines Let the lines be a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

Step 2: Rewrite the equations of the two lines so that their constant terms are positive.

Step 3: The bisector corresponding to the positive sign i.e. (a₁x + b₁y + c₁)/ √ (a₁² + b₁²) = + (a₂x + b₂y + c₂)/ √ (a₂² + b₂²).