The Steepness of Any Road Can Be Captured by One Number Called Its Gradient

The slope or gradient of a line measures how steeply it rises or falls. It is the **change in $y$ for each unit change in $x$**:

$$m=\frac{\text{rise}}{\text{run}}=\frac{y_2-y_1}{x_2-x_1}$$

- Positive slope — the line rises from left to right
- Negative slope — it falls
- Zero slope — it is horizontal

This part covers finding slopes, parallel and perpendicular lines, the form $y=mx+c$, and collinear points.

**From two points use $m=\frac{y_2-y_1}{x_2-x_1}$, from the inclination $\theta$ with the positive x-axis use $m=\tan \theta$, and from an equation rearrange it into the form $y=mx+c$.

From two points.**

$$A(2,3),B(6,11):m=\frac{11-3}{6-2}=\frac{8}{4}=2$$
$$A(-1,4),B(3,-2):m=\frac{-2-4}{3+1}=-\frac{6}{4}=-\frac{3}{2}$$

From the inclination.

$$\theta =45^{\circ }:m=1\theta =60^{\circ }:m=\sqrt{3}\theta =30^{\circ }:m=\frac{1}{\sqrt{3}}\theta =0^{\circ }:m=0$$

From an equation. For $3x-2y+6=0$:

$$2y=3x+6y=\frac{3}{2}x+3m=\frac{3}{2}$$

In general, $ax+by+c=0$ has slope $m=-\frac{a}{b}$.

An everyday example. A ghat road sign showing a gradient of 1 in 10 means the road rises $1$ m for every $10$ m horizontally, a slope of $\frac{1}{10}$.

The boundary case. A vertical line has no defined slope, because $x_2-x_1=0$ and $\tan 90^{\circ }$ is undefined.

**Two lines are parallel when their slopes are equal, $m_1=m_2$, and perpendicular when the product of their slopes is $-1$, $m_1\times m_2=-1$.

Worked example 1.** Are $y=2x+3$ and $4x-2y+7=0$ parallel?

$$4x-2y+7=0\Rightarrow y=2x+\frac{7}{2}{m}_1=m_2=2\text{parallel}$$

Worked example 2. Are $3x+4y=12$ and $4x-3y=5$ perpendicular?

$$m_1=-\frac{3}{4}{m}_2=\frac{4}{3}{m}_1{m}_2=-1\text{perpendicular}$$

Worked example 3. The line through $(2,k)$ and $(4,7)$ is parallel to $y=3x-1$. Find $k$.

$$\frac{7-k}{4-2}=37-k=6k=1$$

Worked example 4. $px+2y=5$ is perpendicular to $y=4x+1$. Find $p$.

$$m=-\frac{p}{2}-\frac{p}{2}\times 4=-1p=\frac{1}{2}$$

An everyday example. Railway tracks are parallel lines with equal slopes, while the crease lines on a cricket pitch meet the edge lines at right angles.

The boundary case. A horizontal and a vertical line are perpendicular, even though the product rule cannot be used, since a vertical line has no slope.

**In $y=mx+c$, $m$ is the gradient and $c$ is the y-intercept, the point $(0,c)$ where the line crosses the y-axis.

Reading an equation.** For $y=-2x+5$: gradient $-2$, so the line **falls $2$ units for every $1$ unit to the right**, and it crosses the y-axis at $(0,5)$.

Worked example 1. Write the line with gradient $\frac{3}{4}$ and y-intercept $-2$.

$$y=\frac{3}{4}x-24y=3x-83x-4y-8=0$$

Worked example 2. A line makes $45^{\circ }$ with the x-axis and cuts the y-axis at $3$: $y=x+3$.

Worked example 3. Find $m$ and $c$ for $2x+5y-10=0$.

$$5y=-2x+10y=-\frac{2}{5}x+2m=-\frac{2}{5},c=2$$

An everyday example. Suppose a taxi charges ₹30 to start and ₹15 per kilometre. The fare is $y=15x+30$: the gradient $15$ is the cost per kilometre and the intercept $30$ is the starting charge.

The substance. **$c$ is the y-intercept, not the x-intercept**; a negative $c$ means the line crosses below the origin.

**Three points $A$, $B$ and $C$ are collinear when the slope of $AB$ equals the slope of $BC$, because the two segments share the point $B$ and so lie on the same line.

Worked example 1.** $A(1,2)$, $B(3,6)$, $C(6,12)$:

$$m_{AB}=\frac{6-2}{3-1}=2m_{BC}=\frac{12-6}{6-3}=2\text{collinear}$$

Worked example 2. $(-2,5)$, $(1,-1)$, $(4,-6)$:

$$m_1=\frac{-1-5}{3}=-2m_2=\frac{-6+1}{3}=-\frac{5}{3}\text{not collinear}$$

Worked example 3. Find $k$ so that $(2,3)$, $(4,k)$ and $(6,-3)$ are collinear.

$$\frac{k-3}{2}=\frac{-3-k}{2}k-3=-3-kk=0$$

An everyday example. A farmer checking that three fence posts stand in a straight line can compare the slopes between them on a plan.

The substance. Equal slopes alone mean parallel; it is the shared point $B$ that makes the lines the same.