Question

The angle between the lines whose intercepts on the axes are $a, - b$and $b,a$ respectively, is
A. ${\tan ^{ - 1}}\left( {\dfrac{{{b^2} - {a^2}}}{2}} \right)$
B. ${\tan ^{ - 1}}\left( {\dfrac{{{a^2} - {b^2}}}{{ab}}} \right)$
C. ${\tan ^{ - 1}}\left( {\dfrac{{{b^2} - {a^2}}}{{2ab}}} \right)$
D. None of these

Answer 1

Hint: In this question the coordinates of axes are given so if we have to find the angle between them, we have to take two lines into consideration and each line will have slope respectively after that we need to find out the angle with help of trigonometry.

Formula Used:
1. In order to find the slope of the given line we can use the following formula:
\[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}},\] where \[{m_1}\] is the slope and \[{y_2},{y_1},{x_2},{x_1}\]are coordinates on\[y\]and \[x\]axes respectively.
2. In order to find the angle between two lines we can use the below mentioned formula:
\[\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|,\] where \[{m_1}\]and \[{m_2}\]are the slopes of the given two lines.

Complete Step-by-step solution
The intercepts given for the first line \[{L_1}\]are $a, - b$so we can say that, this line passes through $(a,0)$ and $(0, - b)$.
\[{m_1} = \dfrac{{ - b - 0}}{{0 - a}}\]Now, we can calculate the slope \[{m_1}\] of this line using formula 1 mentioned above.
\[{m_1} = \dfrac{{ - b}}{{ - a}}\]
\[{m_1} = \dfrac{b}{a}\]
Similarly, the intercepts given for the second line \[{L_2}\] are $b,a$so we can say that, this line passes through $(b,0)$ and $(0,a)$.
Now, we can calculate the slope \[{m_2}\] of this line using formula 1 mentioned above.
\[{m_2} = \dfrac{a-0}{{0 - b}}\]
\[{m_2} = \dfrac{a}{{ - b}}\]
\[{m_2} = \dfrac{{ - a}}{b}\]
As we have calculated both slopes of both the lines. Now using the 2nd formula mentioned above we can calculate the angle between the two lines.
\[\tan \theta = \left| {\dfrac{{\dfrac{b}{a} - \dfrac{{( - a)}}{b}}}{{1 + \dfrac{b}{a} \times \dfrac{{( - a)}}{b}}}} \right|\]
\[\tan \theta = \left| {\dfrac{{\dfrac{b}{a} + \dfrac{a}{b}}}{{1 + ( - 1)}}} \right|\]
\[\tan \theta = \left| {\dfrac{{\dfrac{{{b^2} + {a^2}}}{{ab}}}}{0}} \right|\]
\[\tan \theta = \left| {\dfrac{{{b^2} + {a^2}}}{{ab(0)}}} \right|\]
\[\tan \theta = \left| {\dfrac{{{b^2} + {a^2}}}{0}} \right|\]
\[\tan \theta = \infty \]
\[\tan \theta = \tan \dfrac{\pi }{2}\]
\[\therefore \theta = \dfrac{\pi }{2}\]
Hence the correct option is D.

Note: Considering that this interception point can be marked on a graph gives us a visual idea of the placement of coordinates. Also in this question after solving the value we got which highlights the angle or the value of \[\theta = \dfrac{\pi }{2}\].