The Angle Where Two Roads Meet Can Be Found From Their Slopes Alone

A line's inclination $\theta$ is the angle it makes with the positive x-axis, measured anticlockwise, with $0^{\circ }\le \theta <180^{\circ }$. Its slope is

$$m=\tan \theta$$

One number then captures the direction of the line, and comparing slopes answers questions about parallel lines, right angles, collinear points and the angle between lines without drawing anything.

This part covers finding slopes, the parallel, perpendicular and collinear tests, and the angle between two lines.

**Through $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1\ne x_2$, the slope is $m=\frac{y_2-y_1}{x_2-x_1}$; from the inclination, $m=\tan \theta$.

Worked example 1 — two points.** Through $(3,-2)$ and $(-1,4)$:

$$m=\frac{4-(-2)}{-1-3}=\frac{6}{-4}=-\frac{3}{2}$$

The slope is negative, so the inclination is obtuse.

Worked example 2 — inclination.

$$\theta =60^{\circ }:m=\sqrt{3}\theta =135^{\circ }:m=-1\theta =0^{\circ }:m=0$$

Worked example 3 — inclination from slope. If $m=-\frac{1}{\sqrt{3}}$, then $\tan \theta =-\frac{1}{\sqrt{3}}$, so $\theta =150^{\circ }$.

An everyday example. **A ghat road that rises $1$ m for every $10$ m horizontally** has slope $0.1$.

The boundary case. **A vertical line has $\theta =90^{\circ }$ and no defined slope**, since $x_2-x_1=0$.

**Non-vertical lines are parallel when $m_1=m_2$ and perpendicular when $m_1{m}_2=-1$; three points are collinear when the slope from the first to the second equals the slope from the second to the third.

Worked example 1 — perpendicular.** The line through $(-2,6)$ and $(4,8)$ is perpendicular to the line through $(8,12)$ and $(x,24)$. Find $x$.

$$m_1=\frac{2}{6}=\frac{1}{3}{m}_2=\frac{12}{x-8}\frac{1}{3}\times \frac{12}{x-8}=-1x=4$$

Worked example 2 — parallelogram. For $A(-2,-1)$, $B(4,0)$, $C(3,3)$, $D(-3,2)$:

$$m_{AB}=\frac{1}{6}=m_{DC}{m}_{AD}=\frac{3}{-1}=-3=m_{BC}$$

Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram.

Worked example 3 — collinear. Find $k$ so that $(k,-1)$, $(2,1)$ and $(4,5)$ are collinear.

$$\frac{1-(-1)}{2-k}=\frac{5-1}{4-2}=22=4-2kk=1$$

An everyday example. A town planner checking whether three bus stops lie on one straight road compares the slopes between them on a map grid.

The substance. Two vertical lines are parallel even though their slopes are undefined, and a vertical line is perpendicular to a horizontal one.

**If two non-perpendicular lines have slopes $m_1$ and $m_2$, the acute angle $\theta$ between them satisfies $\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$, and the obtuse angle is $180^{\circ }-\theta$.

Worked example 1.** Slopes $1$ and $\sqrt{3}$ — lines inclined at $45^{\circ }$ and $60^{\circ }$:

$$\tan \theta =\frac{\sqrt{3}-1}{1+\sqrt{3}}=2-\sqrt{3}\theta =15^{\circ }$$

Worked example 2. Slopes $\frac{1}{2}$ and $3$:

$$\tan \theta =\left|\frac{3-\frac{1}{2}}{1+\frac{3}{2}}\right|=\frac{5/2}{5/2}=1\theta =45^{\circ }$$

Worked example 3 — two answers. Find the slope of a line making $45^{\circ }$ with a line of slope $\frac{1}{2}$.

$$\left|\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\right|=1$$

Taking the positive case gives $m=3$; the negative case gives $m=-\frac{1}{3}$. Both lines work.

An everyday example. An architect measuring the angle between two walls on a site plan can compute it from the slopes of the wall lines.

The substance. **When $1+m_1{m}_2=0$ the formula breaks down**, because the lines are perpendicular and the angle is $90^{\circ }$.