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# Show that the equation of the line passing through the origin and making an angle $\theta$ with the line $y=mx+c$ is $\large\frac{y}{x}$$= \large\frac{m \pm \tan \theta}{1 \mp m \tan \theta} Can you answer this question? ## 1 Answer 0 votes Toolbox: • Equation of the line having intercepts a and b on the axes is \large\frac{x}{a}$$+ \large\frac{y}{b}$$=1 Let the equation of the line passing through the origin be y=m_1x \Rightarrow m_1= \large\frac{y}{x} Let then line make an angle \theta with the line y=mx+c \therefore \tan \theta = \bigg| \large\frac{m_1-m}{1+m_1m} \bigg| \Rightarrow \tan \theta = \bigg| \large\frac{\Large\frac{y}{x}-m}{1+ \Large\frac{y}{x}.m} \bigg| \Rightarrow \tan \theta = \large\frac{\bigg( \Large\frac{y}{x}-m \bigg)}{1+ \Large\frac{y}{x}m } or \tan \theta = \large\frac{-\bigg( \large\frac{y}{x}-m \bigg)}{1+\large\frac{y}{x}m} Case (i) If \tan \theta = \large\frac{\bigg( \large\frac{y}{x}-m \bigg)}{1+\large\frac{y}{x}m} \Rightarrow \tan \theta \bigg( 1+ \large\frac{y}{x}$$m \bigg) $$= \large\frac{y}{x}$$-m$
$\Rightarrow \tan \theta + \tan \theta. \large\frac{y}{x}$$m= \large\frac{y}{x}$$-m$
$\Rightarrow m + \tan \theta = \large\frac{y}{x}$$(1- m \: \tan \theta ) \Rightarrow \large\frac{y}{x}$$ = \large\frac{m+ \tan \theta}{1-m \tan \theta}$
Case (ii)
When $\tan \theta = \large\frac{-\bigg( \large\frac{y}{x}-m\bigg)}{1+ \large\frac{y}{x}m}$
$\Rightarrow \tan \theta \bigg( 1+ \large\frac{y}{x}$$m \bigg) = - \bigg( \large\frac{y}{x}$$-m \bigg)$
$\Rightarrow \tan \theta + \tan \theta \large\frac{y}{x}$$m = -\large\frac{y}{x}$$+m$
$\therefore \large\frac{y}{x}$$( 1+ m \tan \theta)=m-\tan \theta \Rightarrow \large\frac{y}{x}$$= \large\frac{m-\tan \theta}{1+m \tan \theta}$
Hence the required line is given by
$\large\frac{y}{x} = \large\frac{m \pm \tan \theta}{1 \mp m \tan \theta}$
Hence proved.