Question

The angle between the lines whos intercepts on the axes are $a,-b$ and $b,-a$ respectively is

• A. ${\tan }^{-1}\frac{\left(a^2-b^2\right)}{ab}$
• B. ${\tan }^{-1}\frac{\left(a^2-b^2\right)}{2ab}$
• C. ${\tan }^{-1}\frac{\left(a^2-b^2\right)}{2}$
• D. none of these

Answer 1

The correct option is B

${\tan }^{-1}\frac{\left(a^2-b^2\right)}{2ab}$

Explanation for correct answer Option(s)

Finding the angle between the lines.

The given first intercepting lines are $b$ and $-a$

Then the line passes through $\left(b,0\right)$ and $\left(0,-a\right)$

Now the slope of the first line is given as,

$$\begin{gathered} m_1=\frac{-\left(-a\right)}{b} \\ {m}_1=\frac{a}{b} \end{gathered}$$

Also consider the given second intercepting lines are $-b$ and $a$

Then the line posses through $\left(a,0\right)$ and $\left(0,-b\right)$

Now the slope of the second line is given as,

$$\begin{gathered} m_2=\frac{-b}{-a} \\ {m}_2=\frac{b}{a} \end{gathered}$$

We know that angle between the two lines is given as,

$\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\theta }\mathbf{=}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Substitute $m_1=\frac{a}{b}$ and $m_2=\frac{b}{a}$ in the above Equation.

$$\begin{gathered} \tan \theta =\left|\frac{{\displaystyle \frac{a}{b}}-{\displaystyle \frac{b}{a}}}{1+\left({\displaystyle \frac{a}{b}}\right)\left({\displaystyle \frac{b}{a}}\right)}\right| \\ \Rightarrow \tan \theta =\left|\frac{{\displaystyle \frac{a^2-b^2}{ab}}}{1+1}\right| \\ \Rightarrow \tan \theta =\left|\frac{a^2-b^2}{2\left(ab\right)}\right| \end{gathered}$$

Now the positive value of $\tan \theta$ is taken:

$$\begin{gathered} \tan \theta =\frac{a^2-b^2}{2ab} \\ \theta ={\tan }^{-1}\left(\frac{a^2-b^2}{2ab}\right) \end{gathered}$$

Hence, the angle between the two lines is ${\tan }^{-1}\frac{\left(a^2-b^2\right)}{2ab}$

Therefore, the correct answer is Option (B)