Show that the equation of the line passing through the origin and making an angle $θ$ with the line y = mx + c is $\frac{y}{x} = \frac{m \pm t a n θ}{1 \mp m t a n θ}$

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Solution

Let $m_{1}$ be the slope of required line which passes through (0, 0).

Then equation of line is y - 0 = $m_{1}$ (x - 0)

$\Rightarrow y = m_{1} x$ . . . .(i)

Now $θ$ is the angle between

y = mx + c and y = $m_{1} x$

$∴ t a n θ = ∣ ∣ \frac{m_{1} - m}{1 + m_{1} 1 m} ∣ ∣ \Rightarrow t a n θ = \pm \frac{m_{1} - m}{1 + m_{1} m}$

$\Rightarrow t a n θ = \frac{m_{1} - m}{1 + m_{1}} o r t a n θ = - \frac{m_{1} - m}{1 + m_{1} m}$

$\Rightarrow t a n θ + m_{1} m t a n θ = m_{1} - m$

or $t a n θ + m_{1} m t a n θ = m - m_{1}$

$\Rightarrow m_{1} (1 - m t a n θ) = m + t a n θ$

or $m_{1} (1 + m t a n θ) = m - t a n θ$

or $t a n θ = - \frac{m_{1} - m}{1 + m_{1} m} \Rightarrow m_{1} = \frac{m + t a n θ}{1 - m t a n θ}$

or $m_{1} = \frac{m - t a n θ}{1 + m t a n θ} \Rightarrow m_{1} = \frac{m \pm t a n θ}{1 \mp m t a n θ}$

Putting value of $m_{1}$ in (i), we have

$y = \pm \frac{m + t a n θ}{1 - m t a n θ} . x$

$\Rightarrow \frac{y}{x} = \pm \frac{m + t a n θ}{1 - m t a n θ} o r \frac{y}{x} = \frac{m \pm t a n θ}{1 \mp t a n θ}$

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