Let $AB:A_1x+B_1y+C_1=0$ and $CD:A_2x+B_2y+C_2=0$ be the two given lines in linear form. Let $B_1$ and $B_2$ be the bisectors of the angles between the two lines. Let $P(h,k)$ be any point on the bisector $B_1$. Draw $PM\perp AB$ and $PN\perp CD$.

\[PM=\pm\left(\frac{A_1h+B_1k+C_1}{\sqrt{A_1^2+B_1^2}}\right)\] \[PN=\pm\left(\frac{A_2h+B_2k+C_2}{\sqrt{A_2^2+B_2^2}}\right)\]

We know that any point on a bisector is equidistant from the two lines. Therefore, \[\pm\left(\frac{A_1h+B_1k+C_1}{\sqrt{A_1^2+B_1^2}}\right)=\pm\left(\frac{A_2h+B_2k+C_2}{\sqrt{A_2^2+B_2^2}}\right)\] \[\left(\frac{A_1h+B_1k+C_1}{\sqrt{A_1^2+B_1^2}}\right)=\pm\left(\frac{A_2h+B_2k+C_2}{\sqrt{A_2^2+B_2^2}}\right)\]

Since these relations are true for any point $(h,k)$ on the bisectors, so the equations of the bisectors are \[\left(\frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}}\right)=\pm\left(\frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}\right)\]

## To identify the bisectors of the angles in which the origin lies

In $AB:A_1x+B_1y+C_1=0$ and $CD:A_2x+B_2y+C_2=0$, let $C_1$ and $C_2$ both be positive. Hence, substitution of the coordinates of the origin in the equations will make both $A_1x+B_1y+C_1$ and $A_2x+B_2y+C_2$ positive. If $Q$ be any point on the bisector containing the origin then the coordinates of $Q$, make $A_1x+B_1y+C_1$ positive because both the points $O$ and $Q$ lie on the same side of the line $AB$ (**From:** The Two Sides of a Line). Similarly, $A_2x+B_2y+C_2$ must also be positive. Hence the equation of the bisector of an angle which contains the origin is \[\frac{A_1h+B_1k+C_1}{\sqrt{A_1^2+B_1^2}}=\frac{A_2h+B_2k+C_2}{\sqrt{A_2^2+B_2^2}}\]

In the same way, the equation of the bisector of the angle in which the origin does not lie is, therefore, \[\frac{A_1h+B_1k+C_1}{\sqrt{A_1^2+B_1^2}}=-\frac{A_2h+B_2k+C_2}{\sqrt{A_2^2+B_2^2}}\] [To apply this, let us not forget that $C_1$ and $C_2$ must both be positive.]

## To identify the bisector of the acute angle and obtuse angle between the lines

Take any bisector $B_1$ and any line $AB$. Let $\theta$ be the angle between the line $AB$ and bisector $B_1$. If $|\tan\theta|<1$, then $\theta<45°$ and $B_1$ is the bisector of an acute angle between the lines $AB$ and $CD$. On the other hand, if $|\tan\theta|>1$, then $\theta>45°$ and $B_1$ bisects the obtuse angle between the lines $AB$ and $CD$.

### Find the equations of the bisectors of the angles between the lines $x-2y=0$ and $2y-11x=6$.

Given lines are, \[x-2y=0\] \[11x-2y+6=0\] The equations of the bisectors of the angles between the lines are given by, \[\frac{x-2y}{\sqrt{1^2+2^2}}=\pm\left(\frac{11x-2y+6}{\sqrt{11^2+2^2}}\right)\] \[\frac{x-2y}{\sqrt{5}}=\pm\frac{11x-2y+6}{5\sqrt{5}}\] \[5x-10y=\pm(11x-2y+6)\] \[\therefore 5x-10y=11x-2y+6\] \[6x+8y+6=0\] \[3x+4y+3=0\] \[\text{and, }5x-10y=-11x+2y-6\] \[16x-12y+6=0\] \[8x-6y+3=0\] Thus, the equations of the bisectors are $3x+4y+3=0$ and $8x-6y+3=0$.

### Find the equations of the bisectors of the angles between the lines $y=x$ and $y=7x+4$. Identify the bisector of the acute angle.

Given lines are, \[x-y=0\] \[7x-y+4=0\] The equations of the bisectors of the angles between the lines are given by, \[\frac{x-y}{\sqrt{1+1}}=\pm\frac{7x-y+4}{\sqrt{7^2+1^2}}\] \[\frac{x-y}{\sqrt{2}}=\pm\frac{7x-y+4}{5\sqrt{2}}\] \[5x-5y=\pm(7x-y+4)\] \[\therefore 5x-5y=7x-y+4\] \[2x+4y+4=0\] \[x+2y+2=0\] \[\text{and, }5x-5y=-7x+y-4\] \[12x-6y+4=0\] \[6x-3y+2=0\] Thus, the equations of the bisectors are $x+2y+2=0$ and $6x-3y+2=0$. Now, to see which is the bisector of an acute angle, we find the angle between on of the bisectors and any one of the given lines.

\[\text{Slope of }x-y=0\text{ }(m_1)=1\] \[\text{Slope of }x+2y+2=0\text{ }m_2=-\frac{1}{2}\] Let $\theta$ be the angle between the line $x-y=0$ and the bisector $x+2y+2=0$. Then, \[\tan\theta=\frac{m_1-m_2}{1+m_1m_2}=\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3>0\] \[\therefore\theta>45°\] Hence, $x+2y+2=0$ is the bisector of obtuse angle. Then, $6x-3y+2=0$ is the bisector of acute angle.

### Find the equations of the bisectors of the angles between the lines $4x-3y+1=0$ and $12x-5y+7=0$ and prove that the bisectors are at right angles to each other. Also identify the bisector of the angle between the lines containing the origin.

Given lines are, \[4x-3y+1=0\] \[12x-5y+7=0\] The equations of the bisectors of the angles between the lines are given by, \[\frac{4x-3y+1}{\sqrt{4^2+3^2}}=\pm\frac{12x-5y+7}{\sqrt{12^2+5^2}}\] \[\frac{4x-3y+1}{5}=\pm\frac{12x-5y+7}{13}\] \[52x-39y+13=\pm(60x-25y+35)\] Now, the equation of the bisector which contains the origin is, \[52x-39y+13=60x-25y+35\] \[8x+14y+22=0\] \[4x+7y+11=0\] and, the equation of the bisector which does not contain the origin is, \[52x-39y+13=-60x+25y-35\] \[112x-64y+48=0\] \[7x-4y+3=0\] Now, \[\text{Slope of }4x+7y+11=0\text{ }(m_1)=-\frac{4}{7}\] \[\text{Slope of }7x-4y+3\text{ }(m_2)=\frac{7}{4}\] \[\therefore m_1×m_2=-\frac{4}{7}×\frac{7}{4}=-1\] Hence, the bisectors are at right angles to each other.