The slope of a line is double of the slope of another line. If tangents of the angle between them is $\frac{1}{3}$ , find the slopes of the other line.

Open in App

Solution

Let $m_{1} and m_{2}$ be the slopes of the given lines.

$∴ m_{2} = 2 m_{1}$

Let $θ$ be the angle between the given lines.

$∴ \tan θ = |\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}| \Rightarrow \frac{1}{3} = |\frac{2 m_{1} - m_{1}}{1 + 2 {m_{1}}^{2}}| = |\frac{m_{1}}{1 + 2 {m_{1}}^{2}}| \Rightarrow \frac{m_{1}}{1 + 2 {m_{1}}^{2}} = \pm \frac{1}{3}$

Taking the positive sign, we get,

$3 m_{1} = 1 + 2 {m_{1}}^{2} \Rightarrow 2 {m_{1}}^{2} - 3 m_{1} + 1 = 0 \Rightarrow (2 m_{1} - 1) (m_{1} - 1) = 0 \Rightarrow m_{1} = \frac{1}{2}, 1$

Taking the negative sign, we get,

$- 3 m_{1} = 1 + 2 {m_{1}}^{2} \Rightarrow 2 {m_{1}}^{2} + 3 m_{1} + 1 = 0 \Rightarrow (2 m_{1} + 1) (m_{1} + 1) = 0 \Rightarrow m_{1} = - \frac{1}{2}, - 1$

Hence, the slopes of the other line are $\pm \frac{1}{2}, \pm 1$ .

Suggest Corrections

Similar questions

\frac{1}{3}

\frac{1}{3}

\frac{1}{3}

Join BYJU'S Learning Program