Pair of Straight Lines
2.0 Angle between the pair of straight lines
2.0 Angle between the pair of straight lines
The acute angle $\theta $ between the pair of straight lines represented by $a{x^2} + 2hxy + b{y^2} = 0$ is given by $$\theta = {\tan ^{ - 1}}\{ \frac{{2\sqrt {({h^2} - ab)} }}{{\left| {a + b} \right|}}\} $$
Proof: As we know that the angle between two lines is calculated using the formulae $$\tan \theta = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$$
Put the values from equation $(4)$ and $(5)$, we get $$\begin{equation} \begin{aligned} \tan \theta = \left| {\frac{{\frac{2}{b}\sqrt {({h^2} - ab)} }}{{1 + \frac{a}{b}}}} \right| = \frac{{2\sqrt {({h^2} - ab)} }}{{a + b}} \\ \theta = {\tan ^{ - 1}}\{ \frac{{2\sqrt {({h^2} - ab)} }}{{\left| {a + b} \right|}}\} \\\end{aligned} \end{equation} $$
Conditions:
- The lines are perpendicular if $\theta = \frac{\pi }{2}$ i.e., $\cot \theta = 0$ $$\begin{equation} \begin{aligned} \frac{{\left| {a + b} \right|}}{{2\sqrt {({h^2} - ab)} }} = 0 \\ a + b = 0 \\{\text{coefficient of }}{x^2} + {\text{coefficient of }}{y^2} = 0 \\\end{aligned} \end{equation} $$
- The lines are coincident, if $\theta = {0^ \circ }{\text{ or }}\pi $ i.e., $\tan \theta = 0$$$\begin{equation} \begin{aligned} \frac{{2\sqrt {({h^2} - ab)} }}{{\left| {a + b} \right|}} = 0 \\ {h^2} - ab = 0 \\ {h^2} = ab \\\end{aligned} \end{equation} $$
Note:
- The pair of lines perpendicular to the lines represented by $a{x^2} + 2hxy + b{y^2} = 0$ and pass through origin can be found out by interchanging the coefficients of ${x^2}$ and ${y^2}$ and change the sign of $xy$ i.e.,$$b{x^2} - 2hxy + a{y^2} = 0$$
Question 1. Find the angle between the pair of straight lines $4{x^2} + 24xy + 11{y^2} = 0$.
Solution: Comparing the given equation of pair of straight line with $$a{x^2} + 2hxy + b{y^2} = 0$$ we get, $a=4$, $b=11$ and $h=12$.
Now, $$\begin{equation} \begin{aligned} \tan \theta = \frac{{2\sqrt {({h^2} - ab)} }}{{\left| {a + b} \right|}} = \frac{{2\sqrt {(144 - 44)} }}{{\left| {4 + 11} \right|}} = \frac{4}{3} \\ \therefore {\text{ }}\theta = {\tan ^{ - 1}}\frac{4}{3} \\\end{aligned} \end{equation} $$ which is the acute angle between the lines.
Question 2. Find the equations to the pair of lines through the origin which are perpendicular to the lines represented by $2{x^2} - 7xy + 3{y^2} = 0$.
Solution: Comparing the given equation of pair of straight line with $$a{x^2} + 2hxy + b{y^2} = 0$$ we get, $a=2$, $b=3$ and $h = - \frac{7}{2}$.
The pair of lines perpendicular to the lines represented by $a{x^2} + 2hxy + b{y^2} = 0$ and pass through origin can be found out by interchanging the coefficients of ${x^2}$ and ${y^2}$ and change the sign of $xy$ i.e.,$$b{x^2} - 2hxy + a{y^2} = 0$$
Therefore, the equation of required line is $$\begin{equation} \begin{aligned} 3{x^2} + 7xy + 2{y^2} = 0 \\ (3x + y)(x + 2y) = 0 \\ 3x + y = 0{\text{ and }}x + 2y = 0 \\\end{aligned} \end{equation} $$
