Chapter 12, Problem 1CLT

To determine

To calculate:Angle between $x-3y=-1$ and $-8x+2y=12$

Answer to Problem 1CLT

Angle between $x-3y=-1$ and $-8x+2y=12$ is ${\tan }^{-1}(\frac{11}{7})$ (in radians).

Angle between $x-3y=-1$ and $-8x+2y=12$ is $57.53°$ (in degrees).

Explanation of Solution

Given information:The two straight lines are $x-3y=-1$ and $-8x+2y=12$

Formula used: $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Calculation:

  $\begin{array}{l}x-3y=-1\\ \Rightarrow -3y=-x-1\\ \Rightarrow y=\frac{1}{3}x+\frac{1}{3}\mathrm{..................}(1)\end{array}$

  $\begin{array}{l}-8x+2y=12\\ \Rightarrow 2y=8x+12\\ \Rightarrow y=4x+\mathrm{6...................}(2)\end{array}$

Let $m_1$ and $m_2$ be the slopes of the straight lines (1) and (2) respectively.

Then, $m_1=\frac{1}{3}$ and $m_2=4$

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

Then,

  $\begin{array}{l}\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|\\ \Rightarrow \tan \theta =\left|\frac{\frac{1}{3}-4}{1+\frac{1}{3}\cdot 4}\right|\\ \Rightarrow \tan \theta =\left|\frac{-\frac{11}{3}}{\frac{7}{3}}\right|\end{array}$

  $\begin{array}{l}\Rightarrow \tan \theta =\left|-\frac{11}{7}\right|\\ \Rightarrow \tan \theta =\frac{11}{7}\\ \Rightarrow \theta ={\tan }^{-1}(\frac{11}{7})\end{array}$

Converting ${\tan }^{-1}(\frac{11}{7})$ to degrees we get $57.53°$ .

Thus, angle between $x-3y=-1$ and $-8x+2y=12$ is ${\tan }^{-1}(\frac{11}{7})$ (in radians) and $57.53°$ (in degrees).