The tangent of angle between the lines
Question:

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

A. $\frac{a^{2}-b^{2}}{a b}$

B. $\frac{b^{2}-a^{2}}{2}$

c. $\frac{b^{2}-a^{2}}{2 a b}$

D. None of these

Solution:

C. $\frac{b^{2}-a^{2}}{2 a b}$

Explanation:

Let the first equation of line having intercepts on the axes $a,-b$ is

$\frac{x}{a}+\frac{y}{-b}=1$

$\Rightarrow \frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}=1$

$\Rightarrow \mathrm{bx}-\mathrm{ay}=\mathrm{ab} \ldots$ (i)

Let the second equation of line having intercepts on the axes $b,-a$ is

$\frac{x}{b}+\frac{y}{-a}=1$

$\Rightarrow \frac{\mathrm{x}}{\mathrm{b}}-\frac{\mathrm{y}}{\mathrm{a}}=1$

$\Rightarrow \mathrm{ax}-\mathrm{by}=\mathrm{ab} \ldots$ (ii)

Now, we find the slope of equation (i)

$b x-a y=a b$

$\Rightarrow a y=b x-a b$

$\Rightarrow y=\frac{b}{a} x-b$

Since, the above equation is in $y=m x+b$ form So, the slope of eq. (i) is

So, the slope of eq. (i) is

$\mathrm{m}_{1}=\frac{\mathrm{b}}{\mathrm{a}}$

Now, we find the slope of equation (ii)

$a x-b y=a b$

$\Rightarrow b y=a x-a b$

$\Rightarrow y=\frac{a}{b} x-a$

Since, the above equation is in $\mathrm{y}=\mathrm{mx}+\mathrm{b}$ form

$a x-b y=a b$

$\Rightarrow \mathrm{by}=\mathrm{ax}-\mathrm{ab}$

$\Rightarrow y=\frac{a}{b} x-a$

Since, the above equation is in $\mathrm{y}=\mathrm{m} \mathrm{x}+\mathrm{b}$ form

So, the slope of equation (i) is

$\mathrm{m}_{1}=\frac{\mathrm{b}}{\mathrm{a}}$

Now, we find the slope of equation (ii)

$a x-b y=a b$

$\Rightarrow \mathrm{by}=\mathrm{ax}-\mathrm{ab}$

$\Rightarrow y=\frac{a}{b} x-a$

Since, the above equation is in $\mathrm{y}=\mathrm{m} \mathrm{x}+\mathrm{b}$ form So, the slope of eq. (i) is

$\mathrm{m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}$

Let $\theta$ be the angle between the given two lines.

$\tan \theta=\left|\frac{m_{1}-m_{1}}{1+m_{1} m_{2}}\right|$

Putting the values of $m_{1}$ and $m_{2}$ in above equation, we get

$\Rightarrow \tan \theta=\left|\frac{\frac{b}{a}-\frac{a}{b}}{1+\left(\frac{b}{a}\right)\left(\frac{a}{b}\right)}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{b^{2}-a^{2}}{a b}}{1+1}\right|$

$\Rightarrow \tan \theta=\left|\frac{b^{2}-a^{2}}{2 a b}\right|$

$\Rightarrow \tan \theta=\frac{b^{2}-a^{2}}{2 a b}$

Hence, the correct option is (c)