Question

Find the angles between each of the following pairs of straight lines:
(i) 3x + y + 12 = 0 and x + 2y − 1 = 0
(ii) 3x − y + 5 = 0 and x − 3y + 1 = 0
(iii) 3x + 4y − 7 = 0 and 4x − 3y + 5 = 0
(iv) x − 4y = 3 and 6x − y = 11
(v) (m² − mn) y = (mn + n²) x + n³ and (mn + m²) y = (mn − n²) x + m³.

Answer 1

(i) The equations of the lines are

3x + y + 12 = 0 ... (1)

x + 2y − 1 = 0 ... (2)

Let $m_1\mathrm{and}{m}_2$ be the slopes of these lines.

$m_1=-3,m_2=-\frac{1}{2}$

Let $\theta$ be the angle between the lines.
Then,

$$\begin{gathered} \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right| \\ =\left|\frac{-3+{\displaystyle \frac{1}{2}}}{1+{\displaystyle \frac{3}{2}}}\right| \\ =1 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{4}\mathrm{or}45° \end{gathered}$$

Hence, the acute angle between the lines is 45$°$

(ii) The equations of the lines are

3x − y + 5 = 0 ... (1)

x − 3y + 1 = 0 ... (2)

Let $m_1\mathrm{and}{m}_2$ be the slopes of these lines.

$m_1=3,m_2=\frac{1}{3}$

Let $\theta$ be the angle between the lines.
Then,

$$\begin{gathered} \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right| \\ =\left|\frac{3-{\displaystyle \frac{1}{3}}}{1+{\displaystyle 1}}\right| \\ =\frac{4}{3} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{4}{3}\right) \end{gathered}$$

Hence, the acute angle between the lines is ${\tan }^{-1}\left(\frac{4}{3}\right)$.

(iii) The equations of the lines are

3x + 4y − 7 = 0 ... (1)

4x − 3y + 5 = 0 ... (2)

Let $m_1\mathrm{and}{m}_2$ be the slopes of these lines.

$m_1=-\frac{3}{4},m_2=\frac{4}{3}$

$$\begin{gathered} \because m_1{m}_2=-\frac{3}{4}\times \frac{4}{3} \\ =-1 \end{gathered}$$

Hence, the given lines are perpendicular.
Therefore, the angle between them is 90^°.

(iv) The equations of the lines are

x − 4y = 3 ... (1)

6x − y = 11 ... (2)

Let $m_1\mathrm{and}{m}_2$ be the slopes of these lines.

$m_1=\frac{1}{4},m_2=6$

Let $\theta$ be the angle between the lines.
Then,

$$\begin{gathered} \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right| \\ =\left|\frac{{\displaystyle \frac{1}{4}}-{\displaystyle 6}}{1+{\displaystyle \frac{3}{2}}}\right| \\ =\frac{23}{10} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{23}{10}\right) \end{gathered}$$

Hence, the acute angle between the lines is ${\tan }^{-1}\left(\frac{23}{10}\right)$

(v) The equations of the lines are

(m² − mn) y = (mn + n²) x + n³ ... (1)

(mn + m²) y = (mn − n²) x + m³ ... (2)

Let $m_1\mathrm{and}{m}_2$ be the slopes of these lines.

$\therefore m_1=\frac{mn+n^2}{m^2-mn},m_2=\frac{mn-n^2}{mn+m^2}$

Let $\theta$ be the angle between the lines.

Then,

$$\begin{gathered} \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right| \\ =\left|\frac{{\displaystyle \frac{mn+n^2}{m^2-mn}-\frac{mn-n^2}{mn+m^2}}}{{\displaystyle 1+}{\displaystyle \frac{mn+n^2}{m^2-mn}}\times {\displaystyle \frac{mn-n^2}{mn+m^2}}}\right| \\ \Rightarrow \tan \theta =\left|\frac{{\displaystyle \left(mn+n^2\right)\left(mn+m^2\right)-\left(mn-n^2\right)\left(m^2-mn\right)}}{\left(m^2-mn\right){\displaystyle \left(mn+m^2\right)+}{\displaystyle \left(mn+n^2\right)\left(mn-n^2\right)}}\right| \\ \Rightarrow \tan \theta =\left|\frac{m^2{n}^2+m^{3}n+m{n}^3+m^2{n}^2-m^{3}n+m^2{n}^2+m^2{n}^2-m{n}^3}{m^{3}n+m^4-m^2{n}^2-m^{3}n+m^2{n}^2-m{n}^3+m{n}^3-n^4}\right| \end{gathered}$$

$\Rightarrow \tan \mathrm{\theta }=\left|\frac{4m^2{n}^2}{{\displaystyle m^4-n^4}}\right|$

Hence, the acute angle between the lines is ${\tan }^{-1}\left(\frac{4m^2{n}^2}{m^4-n^4}\right)$.