# Angle between Two Lines

Let $AB:y=m_1x+c_1$ $CD:y=m_2x+c_2$ be the equations of two lines in slope intercept form. Let $AB$ make an angle $\theta_1$ with the positive x-axis and $CD$ make an angle $\theta_2$ with positive x-axis. $\therefore \text{Slope of AB }(m_1)=\tan\theta_1$ $\text{Slope of CD }(m_2)=\tan\theta_2$ Again, let the angle between two lines $AB$ and $CD$ be $\angle APD=\phi$.

Then, $\theta_2=\theta_1+\phi$ $\phi=\theta_2-\theta_1$ $\tan\phi=\tan(\theta_2-\theta_1)$ $\tan\phi=\frac{\tan\theta_2-\tan\theta_1}{1+\tan\theta_2\tan\theta_1}$ $\tan\phi=\frac{m_2-m_1}{1+m_2m_1}$ $\therefore\tan\phi=-\left(\frac{m_1-m_2}{1+m_1m_2}\right)$ Again, if $\alpha=180°-\phi$ be taken as the angle between two lines, then, $\tan\alpha=\tan(180°-\phi)$ $\tan\alpha=-\tan\phi$ $\therefore\tan\alpha=\frac{m_1-m_2}{1+m_1m_2}$

Hence, if $\theta$ be the angle between two lines, then, $\tan\theta=\pm\left(\frac{m_1-m_2}{1+m_1m_2}\right)$ The positive value of $\tan\theta$ gives the acute angle between the two lines and negative value gives the obtuse angle.

### Condition of Parallelism

The two lines will be parallel to each other if $\theta=0°$. $\therefore\tan 0°=\pm\left(\frac{m_1-m_2}{1+m_1m_2}\right)$ $0=m_1-m_2$ $\therefore m_1=m_2$ Hence, if the slopes are equal, the two lines will be parallel to each other.

### Condition of Perpendicularity

The two lines will be perpendicular to each other if $\theta=90°$. $\therefore\tan 90°=\pm\left(\frac{m_1-m_2}{1+m_1m_2}\right)$ $\cot 90°=\pm\left(\frac{1+m_1m_2}{m_1-m_2}\right)$ $0=\pm\left(\frac{1+m_1m_2}{m_1-m_2}\right)$ $\therefore m_1m_2=-1$ Thus if the product of the slopes is -1, the two lines will be at right angles.

## Angle between Two Lines in Linear Form

Let the equations of the lines be given in the linear form: $AB:A_1x+B_1y+C_1=0$ $CD:A_2x+B_2y+C_2=0$ Then, $\text{Slope of AB }(m_1)=-\frac{A_1}{B_1}$ $\text{Slope of CD }(m_2)=-\frac{A_2}{B_2}$ [From: Reduction of Linear Equation to Slope Intercept Form (Linear Equation $Ax+By+C=0$)]

Hence, the angle between two lines $AB$ and $CD$ is given by $\tan\theta=\pm\left(\frac{-\frac{A_1}{B_1}+\frac{A_2}{B_2}}{1+\frac{A_1A_2}{B_1B_2}}\right)$ $\therefore\tan\theta=\pm\left(\frac{A_1B_2-A_2B_1}{A_1A_2+B_1B_2}\right)$ Now, two lines will be parallel to each other if $\theta=0°$. $\therefore\tan 0°=\pm\left(\frac{A_1B_2-A_2B_1}{A_1A_2+B_1B_2}\right)$ $0=A_1B_2-A_2B_1$ $\therefore\frac{A_1}{A_2}=\frac{B_1}{B_2}\text{ (Parallelism)}$

Again, the two lines will be perpendicular to each other if $\theta=90°$. $\therefore\tan 90°=\pm\left(\frac{A_1B_2-A_2B_1}{A_1A_2+B_1B_2}\right)$ $\cot 90°=\pm\left(\frac{A_1A_2+B_1B_2}{A_1B_2-A_2B_1}\right)$ $0=\pm\left(\frac{A_1A_2+B_1B_2}{A_1B_2-A_2B_1}\right)$ $\therefore A_1A_2+B_1B_2=0\text{ (Perpendicularity)}$

# Equation of any line parallel to the line $Ax+By+C=0$

The given equation of the line is $Ax+By+C=0\text{ __(1)}$ $\therefore\text{Slope of the given line }(m)=-\frac{A}{B}$ Let the given line be parallel to the line $y=mx+c\text{ __(2)}$ $\text{Slope of the line }=m$ Since the lines $\text{(1)}$ and $\text{(2)}$ are parallel, $m=-\frac{A}{B}$ Substituting the value of $m$ in $\text{(2)}$, we have, $y=-\frac{A}{B}x+c$ $\Rightarrow Ax+By-Bc=0$ which is in the form of linear equation $Ax+By+k=0$ where $k=-Bc$.

Thus, to get an equation of a line parallel to the given line, we simply change the constant term of the given equation.

# Equation of any line perpendicular to the line $Ax+By+C=0$

The given equation of the line is $Ax+By+C=0\text{ __(i)}$ $\therefore\text{Slope of the given line }(m)=-\frac{A}{B}$ Let the given line be parallel to the line $y=mx+c\text{ __(ii)}$ $\text{Slope of the line }=m$ Since the lines $\text{(i)}$ and $\text{(ii)}$ are perpendicular, $m\left(-\frac{A}{B}\right)=-1$ $\therefore m=\frac{B}{A}$ Substituting the value of $m$ in $\text{(ii)}$, we have, $y=\frac{B}{A}x+c$ $\Rightarrow Bx-Ay+Ac=0$ which is in the form of linear equation $Bx-Ay+k=0$ where $k=Ac$.

Thus, to get an equation of a line perpendicular to the given line, we interchange the coefficients of $x$ and $y$, change the sign of one of them and also the constant term.

### Find the angle between $2x-3y+7=0$ and $7x+4y-9=0$.

Here, $2x-3y+7=0$ $\therefore\text{Slope }(m_1)=\frac{2}{3}$ $7x+4y-9=0$ $\therefore\text{Slope }(m_2)=-\frac{7}{4}$ Let $\theta$ be the angle between two lines. Then, $\tan\theta=\pm\left(\frac{m_1-m_2}{1+m_1m_2}\right)$ $=\pm\left(\frac{\frac{2}{3}+\frac{7}{4}}{1-\frac{2}{3}.\frac{7}{4}}\right)$ $=\pm\frac{29}{2}$ $\therefore\theta=\tan\left(\pm\frac{29}{2}\right)$

### Determine $k$ so that the line $3x-ky-8=0$ shall make an angle of $45°$ with the line $3x+5y-17=0$.

Given lines are $3x-ky-8=0\text{ __(1)}$ $3x+5y-17=0\text{ __(2)}$ $\text{Slope of equation 1 }(m_1)=\frac{3}{k}$ $\text{Slope of equation 2 }(m_2)=-\frac{3}{5}$ Let $\theta$ be the angle between lines. Then, $\tan\theta=\pm\left(\frac{m_1-m_2}{1+m_1m_2}\right)$ $\tan 45°=\pm\left(\frac{\frac{3}{k}+\frac{3}{5}}{1-\frac{3}{k}.\frac{3}{5}}\right)$ $\pm 1=\frac{15+3k}{5k-9}$ $\therefore\frac{15+3k}{5k-9}=1\text{ or }\frac{15+3k}{5k-9}=-1$ $15+3k=5k-9\text{ or }15+3k=-5k+9$ $24=2k\text{ or }8k=-6$ $\therefore k=12\text{ or }k=-\frac{3}{4}$

### Find the equation of the line through the point $(2,3)$ and parallel to the straight line $4x-3y=10$.

Given line is, $4x-3y-10=0$ $\text{Its Slope }(m)=\frac{4}{3}$ Since the required line is parallel to the given line, so, $\text{Slope of the required line }(m)=\frac{4}{3}$ Thus, equation of the required line is $y-y_1=m(x-x_1)$ $y-y_1=\frac{4}{3}(x-x_1)$ Since the line passes through the point $(2,3)$, $y-3=\frac{4}{3}(x-2)$ $3y-9=4x-8$ $\therefore 4x-3y+1=0$ Alternative Method

Given line is, $4x-3y-10=0$ Any line parallel to the given line is, $4x-3y+k=0$ Since it passes through $(2,3)$, $4×2-3×3+k=0$ $\therefore k=1$ Thus, the required line is, $4x-3y+1=0$

### Find the equation of the line through $(5,4)$ and perpendicular to the line $4x-3y=10$.

Given line is, $4x-3y-10=0$ $\text{Its Slope}=\frac{4}{3}$ Since the required line is perpendicular to the given line, $\text{Slope of the required line }(m)=-\frac{3}{4}$ Thus, the equation of the required line is, $y-y_1=m(x-x_1)$ $y-y_1=-\frac{3}{4}(x-x_1)$ Since it passes through $(5,4)$, $y-4=-\frac{3}{4}(x-5)$ $4y-16=-3x+15$ $\therefore 3x+4y-31=0$ Alternative Method

Given line is, $4x-3y-10=0$ Any line perpendicular to the given line is, $3x+4y+k=0$ Since it passes through $(5,4)$, $3×5+4×4+k=0$ $\therefore k=-31$ Thus, the required line is, $3x+4y-31=0$