Question

The tangent of the angle between the lines whose intercepts on the axes are a, –b and b, –a respectively, is

(a) $\frac{a^2-b^2}{ab}$

(b) $\frac{b^2-a^2}{2}$

(c) $\frac{b^2-a^2}{2ab}$

(d) none of these

Answer 1

Since intercepts of first line are a and −b, this line passes through the points (a, 0) and (0, −b)

$\therefore \mathrm{Its}\mathrm{slope}\mathrm{is}\frac{-b-0}{0-a}=\frac{b}{a}...\left(1\right)$

Now since intercepts of second line are b, − a
i.e. line passes through (b, 0) and (0, – a)

i.e. Its slope is $\frac{-a-0}{0-b}=\frac{a}{b}...\left(2\right)$

Then angle between the above two lines is given by.
$\begin{array}{rcl}\mathrm{tan\theta }& =& \frac{{\displaystyle \frac{b}{a}}-{\displaystyle \frac{a}{b}}}{1+{\displaystyle \frac{a}{b}}\times {\displaystyle \frac{b}{a}}}=\frac{{\displaystyle \frac{b^2-a^2}{ab}}}{1+1}\\ & =& \frac{b^2-a^2}{2ab}\\ & & \end{array}$

Hence, the correct answer is option C.