Question

The angle between the lines y – x + 5 = 0 and $\sqrt {3x} - y + 7 = 0$ is/are
A.15°
B.60°
C.165°
D.75°

Answer 1

Hint: The angle between two lines is the angle between direction vectors of the lines. The angle between lines is given by $ = \tan \theta \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ where $m_1$ and $m_2$ are the slope of lines.

Complete step-by-step answer:
Let the line be
y - x + 5=0 ––––––––– (1)
$\sqrt {3x} - y + 7 = 0$ ––––––– (2)
We know that angle between 2 line can be found by using formula
$\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_2}m1}}} \right|$
Let the slope of line (1) be $m_1$ & slope of line (2) be $m_2$
Calculating $m_1$
From (1)
y – x + 5 = 0
y = x – 5
The above equation is of the form y = mx + c
Where m is the slope
Thus, $m_1$ = 1
Calculating $m_2$
From (2)
$\sqrt 3 x - y + 7 = 0$
$y = \sqrt 3 x + 7$
The above equation is of the form y = mx + c
Where m is the slope
Thus, ${m_2} = \sqrt 3 $
Angle between two lines is given
$\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_2}{m_1}}}} \right|$
Putting values
$\tan \theta = \left| {\dfrac{{1 - \sqrt 3 }}{{1 + \sqrt 3 }}} \right|$
$ = \dfrac{{1 - \sqrt 3 }}{{1 + \sqrt 3 }} \times \dfrac{{1 - \sqrt 3 }}{{1 - \sqrt 3 }}$
$ = \dfrac{{{{\left( {1 - \sqrt 3 } \right)}^2}}}{{1 - 3}}$
$ = \dfrac{{1 + 3 - 2\sqrt 3 }}{{ - 2}}$
$ = \dfrac{{2\left( {2 - \sqrt 3 } \right)}}{{ - 2}}$
$Q = {\tan ^{ - 1}}\left( {2 - \sqrt 3 } \right)$
Q = 15°
Thus, the acute angle between the lines (1) & (2) is θ = 15°
& obtuse angle between these two lines is
φ=180 – θ
= 180° – 15°
= 165°
So, from the above option both A and C options are correct.

Note: There are always two angles between the lines, one acute angle θ & other obtuse angle φ which are in linear pair,
Thus θ + φ = 180°
φ = 180° – θ