Question

The angle between the lines $ 3x - y + 5 = 0 $ , $ x + 3y - 2 = 0 $ is
A. $ \dfrac{\pi }{2} $
B. $ \dfrac{\pi }{4} $
C. $ 0 $
D. $ \dfrac{\pi }{6} $

Answer 1

Hint: Linear equations are defined for the lines of the coordinate system and represent straight lines. These equations are of the first order. The slope of a line tells the steepness of a line, it can be defined as a change in y per unit change in x .To find the angle between two lines we have to first find their slopes.

Complete step-by-step answer:
Equation of a line is - $ y = mx + c $
where $ m $ is the slope of the line and $ c $ is the intercept of the line on the y-axis.
So we first concert the given lines to this form,
 $ 3x - y + 5 = 0 $ can be rearranged as $ y = 3x + 5 $
The slope of line 1 is, $ {m_1} = 3 $
 $ x + 3y - 2 = 0 $ can be rewritten as -
 $
  3y = - x + 2 \\
   \Rightarrow y = \dfrac{{ - 1}}{3}x + \dfrac{2}{3} \;
  $
The slope of the second line is, $ {m_2} = \dfrac{{ - 1}}{3} $
Now the angle between two lines can be found out by the formula –
 $
\Rightarrow \tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\\
= \left| {\dfrac{{3 - (\dfrac{{ - 1}}{3})}}{{1 + 3 \times (\dfrac{{ - 1}}{3})}}} \right|\\
 = \left| {\dfrac{{3 + \dfrac{1}{3}}}{{1 - 1}}} \right| = \left| {\dfrac{{\dfrac{{10}}{3}}}{0}} \right| \\
\Rightarrow \tan \theta = \infty \\
  \therefore \theta = \dfrac{\pi }{2} \;
  $
Thus, the two lines are perpendicular to each other . The angle between them is $ \dfrac{\pi }{2} $ .
So, the correct answer is “Option A”.

Note: The intersection of two lines forms an angle between them. The value of the angle is found out using the slopes of the lines, slope is also called a tangent. To find the slope of a line we first convert it into the slope-intercept form and then put the value of slopes in the suitable formula. The intercept on the y-axis is the distance between the origin and the point at which the line cuts the y-axis.