**Equation of the Bisectors: **The bisectors of the angle between two straight lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 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.

⇒ \(\left| \frac{{{a}_{1}}h+{{b}_{1}}k+{{c}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}}} \right|=\left| \frac{{{a}_{2}}h+{{b}_{2}}k+{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}}} \right|\).

∴ Locus of (h, k) i.e., the equation of the bisectors are \(\frac{\left| {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}} \right|}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\frac{\left| {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}} \right|}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\),

⇒ a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 ⇒ \(\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\).

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 \(\frac{3x-4y+7}{\sqrt{{{3}^{2}}+{{\left( -4 \right)}^{2}}}}=\pm \frac{12x-5y-8}{\sqrt{{{12}^{2}}+{{\left( -5 \right)}^{2}}}}\),

Or, \(\frac{3x-4y+7}{5}=\pm \frac{12x-5y-8}{13}\).

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 \(\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\).

**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. \(\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=+\frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\).