The angle between the lines whose intercepts | vedclass

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(C) The equations of the lines in intercept form $\frac{x}{x_0} + \frac{y}{y_0} = 1$ are:Line $1$: $\frac{x}{a} + \frac{y}{-b} = 1 \implies \frac{x}{a

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$	an^{-1} \frac{b^2 - a^2}{2ab}$

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(C) The equations of the lines in intercept form $\frac{x}{x_0} + \frac{y}{y_0} = 1$ are:Line $1$: $\frac{x}{a} + \frac{y}{-b} = 1 \implies \frac{x}{a} - \frac{y}{b} = 1 \implies y = \frac{b}{a}x - b$. Thus,slope $m_1 = \frac{b}{a}$.Line $2$: $\frac{x}{b} + \frac{y}{-a} = 1 \implies \frac{x}{b} - \frac{y}{a} = 1 \implies y = \frac{a}{b}x - a$. Thus,slope $m_2 = \frac{a}{b}$.The angle $	heta$ between the lines is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.Substituting the values: $	an 	heta = |\frac{\frac{b}{a} - \frac{a}{b}}{1 + (\frac{b}{a})(\frac{a}{b})}| = |\frac{\frac{b^2 - a^2}{ab}}{1 + 1}| = |\frac{b^2 - a^2}{2ab}|$.Therefore,$	heta = 	an^{-1} |\frac{b^2 - a^2}{2ab}|$.

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

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