The slope of a line is double of the slope of another line. If tangent of the angle between them is $\dfrac{1}{3}$, find the slopes of the lines.

Question

Accepted Answer

> [!TLDR]
> Let the slopes be $m$ and $2m$. Solving $\left|\dfrac{m}{1+2m^2}ight| = \dfrac13$ for both signs gives four valid slope pairs: $(1,2)$, $\left(\dfrac12, 1ight)$, $(-1,-2)$, and $\left(-\dfrac12,-1ight)$.

## Setting up the equation

Let the smaller slope be $m$, so the other slope is $2m$ (double of it). If $	heta$ is the angle between the two lines, the standard formula gives
$$
	an	heta = \left|\frac{m_2 - m_1}{1 + m_1m_2}ight| = \left|\frac{2m-m}{1+m(2m)}ight| = \left|\frac{m}{1+2m^2}ight|.
$$
We are told $	an	heta = \dfrac13$, so
$$
\left|\frac{m}{1+2m^2}ight| = \frac13.
$$
Because of the absolute value, this splits into two cases.

## Case 1: $\dfrac{m}{1+2m^2} = \dfrac13$

Cross-multiplying:
$$
3m = 1 + 2m^2 \quad\Longrightarrow\quad 2m^2 - 3m + 1 = 0.
$$
Factoring:
$$
(2m-1)(m-1) = 0 \quad\Longrightarrow\quad m = 1 \ 	ext{or}\ m = \frac12.
$$

## Case 2: $\dfrac{m}{1+2m^2} = -\dfrac13$

Cross-multiplying:
$$
3m = -1 - 2m^2 \quad\Longrightarrow\quad 2m^2 + 3m + 1 = 0.
$$
Factoring:
$$
(2m+1)(m+1) = 0 \quad\Longrightarrow\quad m = -1 \ 	ext{or}\ m = -\frac12.
$$

## Reading off the slope pairs

Recall the two slopes are $m$ and $2m$. Each value of $m$ found above gives one genuine, distinct pair:

| $m$ | Slopes $(m, 2m)$ | …

Slope value	What the line does
$m > 0$	Rises left to right (uphill)
$m < 0$	Falls left to right (downhill)
$m = 0$	Horizontal (flat)
$m$ undefined	Vertical (straight up/down)

Q.The slope of a line is double of the slope of another line. If tangent of the angle between them is $\dfrac{1}{3}$ , find the slopes of the lines.

Concept understanding — Slope Calculation

Slope Calculation — From Intuition to Precision

The Intuition: Rise Over Run

The Precise Definition

What the Number Tells You

A Worked Example

Setting up the equation

Case 1: $\dfrac{m}{1+2m^2} = \dfrac13$

Case 2: $\dfrac{m}{1+2m^2} = -\dfrac13$

Reading off the slope pairs

Q.The slope of a line is double of the slope of another line. If tangent of the angle between them is 13\dfrac{1}{3}31​, find the slopes of the lines.

Setting up the equation

Case 1: m1+2m2=13\dfrac{m}{1+2m^2} = \dfrac131+2m2m​=31​

Case 2: m1+2m2=−13\dfrac{m}{1+2m^2} = -\dfrac131+2m2m​=−31​

Reading off the slope pairs

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Q.The slope of a line is double of the slope of another line. If tangent of the angle between them is $\dfrac{1}{3}$ , find the slopes of the lines.

Case 1: $\dfrac{m}{1+2m^2} = \dfrac13$

Case 2: $\dfrac{m}{1+2m^2} = -\dfrac13$

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