Question

Find the angles between the straight lines 3x + y + 12 = 0 and x + 2y -1 = 0.

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 $\theta $ between two lines of given slope m1 and m2 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 3x + y + 12 = 0,
3x + y + 12 = 0
y = - 3x - 12
By comparison we can see that the slope m1 = -3.

For the line x + 2y - 1 = 0,
x + 2y - 1 = 0
2y = - x + 1
${\text{y}} = - \dfrac{1}{2}{\text{x}} + \dfrac{1}{2}$
By comparison we can see that the slope m2 = -½.

Now, using the given formula we can find the angle between the two lines as-
$tan\theta = \left| {\dfrac{{{{\text{m}}_2} - {{\text{m}}_1}}}{{1 + {{\text{m}}_1}{{\text{m}}_2}}}} \right|$
$tan\theta = \left| {\dfrac{{ - \dfrac{1}{2} - \left( { - 3} \right)}}{{1 + \left( { - \dfrac{1}{2}} \right) \times \left( { - 3} \right)}}} \right|$
$tan\theta = \left| {\dfrac{{\dfrac{5}{2}}}{{1 + \dfrac{3}{2}}}} \right| = \left| 1 \right|$
To eliminate the the modulus sign, we need to add the plus-minus sign as-
$tan\theta = \pm 1$
$tan\theta = 1$
${\theta} = {45^{\text{o}}}$
Also,
$tan\theta = - 1$
${\theta}= {135^{\text{o}}}$
Therefore the acute angle between the lines is 45o and the obtuse angle is 135o. 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.