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Mathematics · Ch 9 — Straight Lines

Angle Between Two Lines

9.2.3

Angle Between Two Lines

Angle Between Two Lines

When we work with just one line in a plane, we talk about its slope and inclination. But the moment we consider two lines, a natural question arises: what is the angle between them? Two lines in a plane can either intersect or be parallel. For intersecting lines, we need a way to find the angle between them using their slopes — and that is what this section is about.

Setting Up the Problem

Consider two non-vertical lines L1L_1 and L2L_2 with slopes m1m_1 and m2m_2 respectively. Let their inclinations (angles with the positive x-axis) be α1\alpha_1 and α2\alpha_2. From the definition of slope, we have:

m1=tan⁡α1andm2=tan⁡α2m_1 = \tan \alpha_1 \quad \text{and} \quad m_2 = \tan \alpha_2

When two lines intersect, they form two pairs of vertically opposite angles. The sum of any two adjacent angles is 180∘180^\circ. Let θ\theta and ϕ\phi be the two adjacent angles between L1L_1 and L2L_2, as shown in the figure.

From the geometry of the situation, one of these angles equals the difference of the inclinations:

θ=α2−α1\theta = \alpha_2 - \alpha_1

This holds provided α1,α2≠90∘\alpha_1, \alpha_2 \neq 90^\circ (so that the slopes are defined). Using the tangent subtraction formula:

tan⁡θ=tan⁡(α2−α1)=tan⁡α2−tan⁡α11+tan⁡α1tan⁡α2=m2−m11+m1m2\tan \theta = \tan(\alpha_2 - \alpha_1) = \frac{\tan \alpha_2 - \tan \alpha_1}{1 + \tan \alpha_1 \tan \alpha_2} = \frac{m_2 - m_1}{1 + m_1 m_2}

This expression is valid as long as 1+m1m2≠01 + m_1 m_2 \neq 0.

The other adjacent angle ϕ\phi is supplementary to θ\theta:

ϕ=180∘−θ\phi = 180^\circ - \theta

Therefore:

tan⁡ϕ=tan⁡(180∘−θ)=−tan⁡θ=−m2−m11+m1m2=m1−m21+m1m2\tan \phi = \tan(180^\circ - \theta) = -\tan \theta = -\frac{m_2 - m_1}{1 + m_1 m_2} = \frac{m_1 - m_2}{1 + m_1 m_2}

Watch out

The formula tan⁡θ=m2−m11+m1m2\tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2} gives the tangent of one of the two adjacent angles. It could be acute or obtuse depending on the sign of the expression. Do not assume it always gives the acute angle.

Two Cases Arise

The sign of the expression m2−m11+m1m2\frac{m_2 - m_1}{1 + m_1 m_2} determines which of θ\theta and ϕ\phi is acute and which is obtuse.

Case I: If m2−m11+m1m2\frac{m_2 - m_1}{1 + m_1 m_2} is positive, then tan⁡θ\tan \theta is positive and tan⁡ϕ\tan \phi is negative. This means θ\theta is acute (since tan⁡\tan positive for acute angles) and ϕ\phi is obtuse.

Case II: If m2−m11+m1m2\frac{m_2 - m_1}{1 + m_1 m_2} is negative, then tan⁡θ\tan \theta is negative and tan⁡ϕ\tan \phi is positive. This means θ\theta is obtuse and ϕ\phi is acute.

Important

The acute angle θ\theta between two lines with slopes m1m_1 and m2m_2 is always given by:

tan⁡θ=∣m2−m11+m1m2∣\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|

The absolute value ensures we get the acute angle. The obtuse angle ϕ\phi can then be found from ϕ=180∘−θ\phi = 180^\circ - \theta.

The Formula for Acute Angle

The textbook presents the acute angle formula as:

tan⁡θ=∣m2−m11+m1m2∣,where 1+m1m2≠0\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|, \quad \text{where } 1 + m_1 m_2 \neq 0

This formula gives the acute angle between any two non-vertical lines. If 1+m1m2=01 + m_1 m_2 = 0, the lines are perpendicular (since m1m2=−1m_1 m_2 = -1), and the angle between them is 90∘90^\circ.

Worked Example 2: Finding the Slope of the Other Line

Problem: The angle between two lines is π4\frac{\pi}{4} and the slope of one line is 12\frac{1}{2}. Find the slope of the other line.

Solution:

Let m1=12m_1 = \frac{1}{2} and let m2=mm_2 = m be the unknown slope. The acute angle θ=π4\theta = \frac{\pi}{4}.

Using the formula:

tan⁡π4=∣m−121+12m∣\tan \frac{\pi}{4} = \left| \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} \right|

Since tan⁡π4=1\tan \frac{\pi}{4} = 1, we have:

∣m−121+m2∣=1\left| \frac{m - \frac{1}{2}}{1 + \frac{m}{2}} \right| = 1

This gives two possibilities (removing the absolute value):

m−121+m2=1orm−121+m2=−1\frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = 1 \quad \text{or} \quad \frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = -1

Case 1: m−121+m2=1\frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = 1

m−12=1+m2m - \frac{1}{2} = 1 + \frac{m}{2}

m−m2=1+12m - \frac{m}{2} = 1 + \frac{1}{2}

m2=32\frac{m}{2} = \frac{3}{2}

m=3m = 3

Case 2: m−121+m2=−1\frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = -1

m−12=−1−m2m - \frac{1}{2} = -1 - \frac{m}{2}

m+m2=−1+12m + \frac{m}{2} = -1 + \frac{1}{2}

3m2=−12\frac{3m}{2} = -\frac{1}{2}

m=−13m = -\frac{1}{3}

Therefore, the slope of the other line is either 33 or −13-\frac{1}{3}.

Note

Two answers arise because the given angle π4\frac{\pi}{4} could be the acute angle between the lines in two different configurations. The figure in the textbook (Fig 9.7) shows this geometrically — the line with slope 12\frac{1}{2} can make an angle of 45∘45^\circ with two different lines, one steeper and one shallower. …

Figure 9.6Angle between two lines
Fig. 9.6 — Angle between two lines

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Fig 9.6 is a simple coordinate-plane sketch with two non-vertical lines, L₁ and L₂, crossing each other above the x-axis. L₁ is the shallower line (smaller slope), L₂ the steeper one (larger slope). Each line meets the x-axis at a distinct point, and at those crossing points the angle each line makes with the positive x-axis is marked: α₁ for L₁, α₂ for L₂. Both α₁ and α₂ are measured anticlockwise from the positive x-axis, so α₂ > α₁.

At the intersection point of L₁ and L₂, two adjacent angles are labelled: θ (the acute angle) and φ (the obtuse angle). The figure makes clear that θ = α₂ − α₁, because the exterior angle of a triangle equals the sum of the opposite interior angles — or, more directly, because the angle between the two lines is simply the difference of their inclinations.

The physical idea is simple: the steepness of a line is captured by its slope, and the angle between two lines can be expressed purely in terms of their slopes, without needing to draw the lines or measure angles with a protractor. The figure anchors the derivation that follows.

tan⁡θ=m2−m11+m1m2\tan\theta = \frac{m_2 - m_1}{1 + m_1 m_2}

where θ\theta is the acute angle between the lines, m1=tan⁡α1m_1 = \tan\alpha_1, m2=tan⁡α2m_2 = \tan\alpha_2, and 1+m1m2≠01 + m_1 m_2 \neq 0.

The obtuse angle ϕ\phi is 180∘−θ180^\circ - \theta, so tan⁡ϕ=−tan⁡θ\tan\phi = -\tan\theta.

The figure also illustrates why two answers appear when solving for an unknown slope: depending on which line is taken as L₁ and which as L₂, the acute angle formula gives either the slope of the steeper line or the shallower one. In Example 2, the two possible slopes (3 and −13-\frac{1}{3}) correspond to the two lines that make a 45∘45^\circ angle with a given line of slope 12\frac{1}{2} — one on each side of it. …

Figure 9.7A line L and two lines at 45° to it
Fig. 9.7 — A line L and two lines at 45° to it

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Fig. 9.7 is a geometric demonstration of why the angle-between-lines formula gives two possible slopes for the second line when the angle is fixed. The figure shows three lines that all pass through a single point (they are concurrent). One line, labelled L, has slope 12\frac{1}{2}. A second line, labelled L₂, has slope 33. A third line, labelled L₁, has slope −13-\frac{1}{3}. Between L and L₂, and between L and L₁, a small 45∘45^\circ arc is drawn. An '8' tick appears on the x‑axis (likely a coordinate label or a distance mark, though its exact meaning is not elaborated in the text).

The physical idea is this: if you know the slope of one line (m1=12m_1 = \frac{1}{2}) and the acute angle between it and another line (θ=45∘\theta = 45^\circ), there are two distinct lines that satisfy that condition — one on each side of the given line. The line L₂ makes a 45∘45^\circ angle with L when measured in one direction; the line L₁ makes the same 45∘45^\circ angle when measured in the opposite direction. Their slopes are different because the angle is measured from L to each line in opposite senses.

The textbook develops this with the formula for the acute angle θ\theta between two lines of slopes m1m_1 and m2m_2:

tan⁡θ=∣m2−m11+m1m2∣\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|

Here m1m_1 is the known slope (12\frac{1}{2}), θ=π4\theta = \frac{\pi}{4} (45∘45^\circ), and m2m_2 is the unknown slope. Removing the absolute value gives two equations:

m2−121+12m2=1orm2−121+12m2=−1\frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} = 1 \quad \text{or} \quad \frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} = -1

Solving the first yields m2=3m_2 = 3 (line L₂). Solving the second yields m2=−13m_2 = -\frac{1}{3} (line L₁). The figure makes this concrete: both lines L₂ and L₁ are at 45∘45^\circ to L, but on opposite sides, so their slopes are not the same. …