Browse Questions

Choose the correct answer from the given four options. The tangent of angle between the lines whose intercepts on the axes are a, – b and b, – a, respectively, is

$\begin {array} {1 1} (A)\;\large\frac{a^2-b^2}{ab} & \quad (B)\;\large\frac{b^2-a^2}{2} \\ (C)\;\large\frac{b^2-a^2}{2ab} & \quad (D)\;\text{None of these} \end {array}$

Toolbox:
• Slope of a line $'m' = -\large\frac{coefficient \: of \: x}{coefficient \: of \: y}$
• Angle between two lines whose slopes are $m_1$ and $m_2$ is $\tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Step 1:
Equation of the given lines are
$\large\frac{x}{a}$$+\large\frac{y}{-b}$$=1$
$\Rightarrow -bx+ay=-ab$----------(1) and
$\large\frac{x}{b}$$+\large\frac{y}{-a}$$=1$
$\Rightarrow -ax+by=-ab$--------(2)
Slope of the line (1) is $m_1=\large\frac{b}{a}$
Slope of the line (2) is $m_2=\large\frac{a}{b}$
$\therefore \tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Now substituting the values we get,
$\tan \theta = \bigg| \large\frac{\Large\frac{b}{a}-\Large\frac{a}{b}}{1+\Large\frac{b}{a} \times \Large\frac{a}{b}} \bigg|$
$= \large\frac{b^2-a^2}{2ab}$
Hence c is the correct option.