Question

Find the angle between the straight lines $x - 4y = 3$ and $6x - y = 11$.

Answer 1

Hint: The slope of a line is defined as the change in y-coordinates with respect to change in x-coordinates. The angle between two lines $x - 4y = 3$ and $6x - y = 11$ of given slope $m_1$ and $m_2$ is given by the formula-
$tan\theta = \left| {\dfrac{{{{\text{m}}_2} - {{\text{m}}_1}}}{{1 + {{\text{m}}_1}{{\text{m}}_2}}}} \right|$

Complete step-by-step answer:

To find the slope of a line, we will convert it into the slope-intercept form, which is given by-
$y = mx + c$ where m is the slope of the line.
We can compare the coefficient of x to find the slope of the line.

For the line $x - 4y = 3$,
$x - 4y = 3$
$4y = x - 3$
${\text{y}} = \dfrac{1}{4}{\text{x}} - \dfrac{3}{4}$
By comparison we can see that the slope $m_1 = \dfrac{1}{4}$.

For the line $6x - y = 11$,
$6x - y = 11$
$y = 6x - 11$
By comparison we can see that the slope $m_2 = 6$.

Now, using the given formula we can find the angle between the two lines as-
$tan\theta = \left| {\dfrac{{{{\text{m}}_2} - {{\text{m}}_1}}}{{1 + {{\text{m}}_1}{{\text{m}}_2}}}} \right|$
$tan\theta = \left| {\dfrac{{6 - \dfrac{1}{4}}}{{1 + \dfrac{6}{4}}}} \right| = \left| {\dfrac{{\dfrac{{23}}{4}}}{{\dfrac{{10}}{4}}}} \right| = \left| {\dfrac{{23}}{{10}}} \right|$
$tan\theta = \left| {2.3} \right|$
To eliminate the the modulus sign, we need to add the plus-minus sign as-
$tan\theta = \pm 2.3$
$tan\theta = 2.3$
${{\theta }} = {\tan ^{ - 1}}2.3$
${{\theta }} \approx 2.3$
Also,
$\begin{aligned} &\tan \theta = - 2.3 \\\ &\theta = {113.5^o} \\\ \end{aligned}$

Therefore the acute angle between the lines is $66.5^o$ and the obtuse angle is $113.5^o$. This is the required answer.

Note: Students often forget to consider both the cases while finding the angle. We should remember that whenever we eliminate the modulus sign, then we need to replace it by the plus-minus sign, hence we get two cases and two answers. In this case, the acute angle between the lines is $66.5^o$ and the obtuse angle is $113.5^o$.