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

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Solution

Step 1: Finding equation of line that intercepts $x$ -axis and $y$ -axis at $a$ and $- b$ respectively
Equation of a line is given as,
$y = m x + c$
where $m$ and $c$ are the slope and $y$ -intercept of the line.
Given, line intercepts $x$ -axis at $a$ , that means, $y = 0$ when $x = a$
$\Rightarrow 0 = m_{1} \cdot a + c_{1} \Rightarrow m_{1} = - \frac{c_{1}}{a}$
where $m_{1}$ and $c_{1}$ are the slope and $y$ -intercept of the line.
Also given, line intercepts $y$ -axis at $- b$ , that means, $y = a$ when $x = 0$
$\Rightarrow - b = m_{1} \cdot 0 + c_{1} \Rightarrow c_{1} = - b$
But,
$m_{1} = - \frac{c_{1}}{a} \Rightarrow m_{1} = \frac{b}{a}$
Step 2: Finding equation of line that intercepts $x$ -axis and $y$ -axis at $b$ and $- a$ respectively
Given, line intercepts $x$ -axis at $b$ , that means, $y = 0$ when $x = b$
$\Rightarrow 0 = m_{2} \cdot b + c_{2} \Rightarrow m_{2} = - \frac{c_{2}}{b}$
where $m_{2}$ and $c_{2}$ are the slope and $y$ -intercept of the line.
Also given, line intercepts $y$ -axis at $- a$ , that means, $y = b$ when $x = 0$
$\Rightarrow - a = m_{2} \cdot 0 + c_{2} \Rightarrow c_{2} = - a$
But,
$m_{2} = - \frac{c_{2}}{a} \Rightarrow m_{2} = \frac{a}{b}$
Step 3: Finding the tangent of the angle between the lines
Let $θ$ be the angle between the two lines.
Tangent of angle between the two lines is given as,
$\tan (θ) = |\frac{m_{2} - m_{1}}{1 + m_{2} m_{1}}| \Rightarrow \tan (θ) = |\frac{\frac{a}{b} - \frac{b}{a}}{1 + \frac{a}{b} \cdot \frac{b}{a}}| \Rightarrow \tan (θ) = |\frac{\frac{a^{2} - b^{2}}{ab}}{2}| \Rightarrow \tan (θ) = \frac{a^{2} - b^{2}}{2 ab}$
Therefore, the tangent of the angle between the lines, $\tan (θ) = \frac{a^{2} - b^{2}}{2 ab}$

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