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 $\angle APD=\phi$.

Angle between Two Lines

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, \[3x+4y-31=0\]


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