# Angle Bisectors of Angles Formed by Intersecting Lines

**\(\textbf{Art 9 :} \qquad\boxed{\text{Angle Bisectors}}\)**

Consider two straight lines *L*_{1} and *L*_{2} with the equations

\[\begin{align}{L_1}\,\,\,\,:\,\,\,{a_1}x + {b_1}y + {c_1} = 0\\{L_2}\,\,\,\,:\,\,\,{a_2}x + {b_2}y + {c_2} = 0\end{align}\]

We intend to find the angle bisector formed at the intersection point *P* of *L*_{1} and *L*_{2}. Note that there will be two such angle bisectors

To write down the equations of the two angle bisectors, we first modify the equations of *L*_{1} and *L*_{2} so that *c*_{1} and *c*_{2} are say, both negative in sign. This can always be done. Why this is done will soon become clear.

We first write down the equation of *A*_{1}, the angle bisector of the angle in which the origin lies.

By virtue of being an angle bisector, if any point \(P(x',y')\) lies on *A*_{1}, the distance of *P *from *L*_{1} and *L*_{2} must be equal. Using the perpendicular distance formula of Art -7, we have

\[\begin{align}&\Rightarrow\qquad \left| {\frac{{{a_1}x' + {b_1}y' + {c_1}}}{{\sqrt {a_1^{\,\,2} + b_1^{\,\,2}} }}} \right| = \left| {\frac{{{a_2}x' + {b_2}y' + {c_2}}}{{\sqrt {a_2^{\,\,2} + b_2^{\,\,2}} }}} \right|\\&\Rightarrow\qquad\frac{{{a_1}x' + {b_1}y' + {c_1}}}{{\sqrt {a_1^{\,\,2} + b_1^{\,\,2}} }} = \pm \frac{{{a_2}x' + {b_2}y' + {c_2}}}{{\sqrt {a_2^{\,\,2} + b_2^{\,\,2}} }} \qquad \qquad \qquad \qquad \dots \rm{(1)}\end{align}\]

Which sign should we select, “+” or “–”, for the bisector of the angle containing the origin?

Since *P* and origin lie on the same side of \( {L_1,}\,{a_1}x' + {b_1}y' + {c_1}\) and *c*_{1} **must be** of the same sign by Art - 6. Similarly, \({a_2}x' + {b_2}y' + {c_2}\) and *c*_{2} must be of the same sign. But since we have already arranged *c*_{1} and *c*_{2} to be of the same sign (both negative), we must have \(\left( {{a_1}x' + {b_1}y' + {c_1}} \right)\) and \(\left( {{a_2}x' + {b_2}y' + {c_2}} \right)\) also of the same sign.

Thus, it follows from (1) that to write the equation of the angle bisector of the angle containing the origin, we must select the “+” sign since \(\left( {{a_1}x' + {b_1}y' + {c_1}} \right)\) and \(\left( {{a_2}x' + {b_2}y' + {c_2}} \right)\) are of the same sign. The “–” sign gives the angle bisector of the angle not containing the origin, i.e., the equation of *A*_{2}.

To summarize, we first arrange the equations of *L*_{1 }and *L*_{2} so that *c*_{1} and *c*_{2} are both of the same sign. Subsequently, using the property of any angle bisector, we obtain

\[\begin{align}&\boxed{{\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}} }}}}\qquad \qquad: \qquad \qquad \begin{array}{l}{\textbf{Angle bisector of angle}}\\{\textbf{contaning the origin}}\end{array}\\\\& \qquad \qquad \text {and}\\\\

&\boxed{{\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}} }}}}\qquad \qquad: \qquad \qquad \begin{array}{l}{\textbf{Angle bisector of angle not}}\\{\textbf{contaning the origin}}\end{array}\end{align}\]

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