Question

Angle between the lines $\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{a}-\dfrac{y}{b}=1$ is
A. $2 \tan ^{-1}\left(\dfrac{b}{a}\right)$
B. $\tan ^{-1}\left(\dfrac{2 a b}{a^{2}-b^{2}}\right)$
C. $\tan ^{-1}\left(\dfrac{a^{2}-b^{2}}{a^{2}+b^{2}}\right)$
D. None of these

Answer 1

Hint: Here, in the given question, we are given two lines, $\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{a}-\dfrac{y}{b}=1$ and we need to find the angle between these two lines. To find the angle between the lines, we will first find the slopes of the lines between which the angle needs to be determined.

Formula used:
we will find the angle between the lines using $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$ formula.

Complete step by step solution:
We have, $\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{a}-\dfrac{y}{b}=1$
The equation of straight line is given by $y=mx+c$ where $m$ is the slope of the line and $c$ is the intercept. So, we will first find the slope of both the lines
Let the slope of line $\dfrac{x}{a}+\dfrac{y}{b}=1$ be ${{m}_{1}}$
$\Rightarrow \dfrac{bx+ay}{ab}=1$
On cross multiplication, we get
$\Rightarrow bx+ay=ab$
$\Rightarrow ay=-bx+ab$
Divide the above-written equation by $a$
$\Rightarrow y=-\dfrac{b}{a}x+b$
$\therefore {{m}_{1}}=-\dfrac{b}{a}$
Let the slope of line $\dfrac{x}{a}-\dfrac{y}{b}=1$ be ${{m}_{2}}$
$\Rightarrow \dfrac{bx-ay}{ab}=1$
On cross multiplication, we get
$\Rightarrow bx-ay=ab$
$\Rightarrow -ay=-bx+ab$
Divide the above-written equation by $-a$
$\Rightarrow y=\dfrac{b}{a}x-b$
$\therefore {{m}_{2}}=\dfrac{b}{a}$
We know that angle between the lines is given by $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$. Therefore, on substituting the vales, we get
$\Rightarrow \tan \theta =\left| \dfrac{\left( -\dfrac{b}{a} \right)-\left( \dfrac{b}{a} \right)}{1+\left( -\dfrac{b}{a} \right)\left( \dfrac{b}{a} \right)} \right|$
$\Rightarrow \tan \theta =\left| \dfrac{-\dfrac{b}{a}-\dfrac{b}{a}}{1+\left( -\dfrac{{{b}^{2}}}{{{a}^{2}}} \right)} \right|$
On taking L.C.M., we get
$\Rightarrow \tan \theta =\left| \dfrac{\dfrac{-b-b}{a}}{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}} \right|$
$\Rightarrow \tan \theta =\left| \dfrac{-2b}{a}\times \dfrac{{{a}^{2}}}{{{a}^{2}}-{{b}^{2}}} \right|$
On canceling the common terms, we get
$\Rightarrow \tan \theta =\left| \dfrac{-2ba}{{{a}^{2}}-{{b}^{2}}} \right|$
$\Rightarrow \theta ={{\tan }^{-1}}\left| \dfrac{-2ba}{{{a}^{2}}-{{b}^{2}}} \right|$
We know that angle can’t be negative, so we get
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{2ba}{{{a}^{2}}-{{b}^{2}}} \right)$
Hence, the angle between the lines $\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{a}-\dfrac{y}{b}=1$ is ${{\tan }^{-1}}\left( \dfrac{2ab}{{{a}^{2}}-{{b}^{2}}} \right)$.

So the correct answer is option B.

Note: In order to solve these types of questions, one must be familiar with the formulas such as the general equation of the line i.e., $y=mx+c$ and the formula of angle between two lines, which is $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$. Don’t get confuse while substituting the values in $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$ because there is a chance of mistake of a minus sign. If you miss the negative sign the result may get differ so take care of the calculations as to be sure of your final answer.