Question

Find the acute angle between the lines $2x-y+3=0$ and $x-3y+2=0$?

Answer 1

To find the acute angle between the two lines we are going to use the following formula: $\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}{{m}_{1}}}$. In this formula, ${{m}_{1}}\And {{m}_{2}}$ are the slopes of the two straight lines. Let us assume the slope of straight line $2x-y+3=0$ is ${{m}_{1}}$ and slope of the straight line $x-3y+2=0$ is ${{m}_{2}}$. Substitute these values of ${{m}_{1}}\And {{m}_{2}}$ in the above formula.

Complete step by step solution:
The two straight lines given in the above problem of which we have to find the acute angle is as follows:
$2x-y+3=0$ and $x-3y+2=0$
Let us name the slope of straight line $2x-y+3=0$ as ${{m}_{1}}$ and slope of the straight line $x-3y+2=0$ as ${{m}_{2}}$.
Now, to find the angle between these two straight lines we are going to use the following formula:
$\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}{{m}_{1}}}$ ………….. (1)
Now, we are going to find the values of ${{m}_{1}}\And {{m}_{2}}$ of the straight lines $2x-y+3=0$ and $x-3y+2=0$ by using the following formula of slope for $ax+by+c=0$:
$m=-\dfrac{a}{b}$
Writing the slope for $2x-y+3=0$ which is equal to ${{m}_{1}}$ as follows:
${{m}_{1}}=-\dfrac{2}{-1}$
Negative sign will get cancelled out from the numerator and the denominator and we get,
${{m}_{1}}=2$
Now, writing the slope for $x-3y+2=0$ which is equal to ${{m}_{2}}$ as follows:
${{m}_{2}}=-\dfrac{1}{-3}$
Negative sign will get cancelled out from the numerator and the denominator and we get,
${{m}_{2}}=\dfrac{1}{3}$
Substituting the above values of ${{m}_{1}}\And {{m}_{2}}$ in eq. (1) we get,
$\begin{aligned} & \tan \theta =\dfrac{\dfrac{1}{3}-2}{1+\dfrac{1}{3}\left( 2 \right)} \\\ & \Rightarrow \tan \theta =\dfrac{\dfrac{1-2\left( 3 \right)}{3}}{\dfrac{3+2}{3}} \\\ & \Rightarrow \tan \theta =\dfrac{\dfrac{1-6}{3}}{\dfrac{3+2}{3}} \\\ & \Rightarrow \tan \theta =\dfrac{\dfrac{-5}{3}}{\dfrac{5}{3}} \\\ \end{aligned}$
In the above equation, 3 will get cancelled out from the numerator and the denominator and we get,
$\begin{aligned} & \tan \theta =\dfrac{-5}{5} \\\ & \Rightarrow \tan \theta =-1 \\\ & \Rightarrow \theta =\pi -\dfrac{\pi }{4} \\\ \end{aligned}$
Now, in the above question, we are asked to find the acute angle between the two straight lines and the angle which we get from the above equation is the obtuse angle so we have to subtract the above angle from $\pi$ to get the acute angle.
Subtracting $\pi$ from $\theta$ we get,
$\begin{aligned} & \theta =\pi -\left( \pi -\dfrac{\pi }{4} \right) \\\ & \Rightarrow \theta =\pi -\pi +\dfrac{\pi }{4} \\\ & \Rightarrow \theta =\dfrac{\pi }{4} \\\ \end{aligned}$

Hence, the acute angle between two straight lines is $\dfrac{\pi }{4}$.

Note: The mistake that could be possible in the above problem is that you might forget to convert the obtuse angle which we have found in the above solution to acute angle so make sure you won’t make this mistake. Sometimes, even examiners are very smart and they give you the options in which you can see that obtuse angle and mark it as the correct answer so make sure you won’t make this mistake.