The general equation of a line and its gradient – Linear and Quadratic Functions (Senior 3 Mathematics) | Ganzaa

Linear and Quadratic Functions

Lessons

Senior 3 · Mathematics

Linear and Quadratic Functions

5 lessons

Lesson 1 of 5

The general equation of a line and its gradient

25 min

How steep is the road?

Rwanda's roads climb and fall across the hills, and anyone who has pushed a loaded bicycle uphill knows that some stretches are much steeper than others. Mathematics measures steepness with one number, the gradient: how far a line rises for each unit it runs across.

In S1 you drew lines such as $y = 2 x + 1$ and read off their gradient and intercepts. This lesson makes that exact: the gradient between any two points, and the gradient and intercepts of a line written in its general form $a x + b y + c = 0$ .

Definition

Gradient (slope) of the line through $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ , with $x_1 \ne x_2$ :

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

Notation: the gradient is usually called $m$ . The change in $y$ is also written $\Delta y$ ("delta $y$ "), so $m = \frac{\Delta y}{\Delta x}$ .

Example: for $A (1, 1)$ and $B (5, 4)$ , $m = \frac{4 - 1}{5 - 1} = \frac{3}{4}$ : the line rises 3 units for every 4 units across.

The gradient is the rise divided by the run.

Important

What the sign of the gradient tells you

Gradient	The line, read from left to right	Example
$m > 0$	rises	$y = x + 1$
$m < 0$	falls	$y = -\frac{1}{2}x + 3$
$m = 0$	is horizontal	$y = - 2$
undefined	is vertical (the run is 0, and you cannot divide by 0)	$x = 4$

The bigger the size $∣ m ∣$ , the steeper the line.

Definition

Linear function: a function of the form $f (x) = m x + c$ . Its graph is a straight line with gradient $m$ that cuts the $y$ -axis at $(0, c)$ .

The $y$ -intercept is found by putting $x = 0$ .
The $x$ -intercept (the root of $f$ ) is found by putting $y = 0$ : $x = -\frac{c}{m}$ when $m \ne 0$ .

Example: $f (x) = 2 x - 6$ has gradient 2, $y$ -intercept $(0, - 6)$ and $x$ -intercept $(3, 0)$ .

Definition

General Cartesian equation of a line: an equation of the form

a x + b y + c = 0,

where $a$ , $b$ and $c$ are real numbers and $a$ and $b$ are not both zero. Every straight line, including vertical ones, can be written this way.

Example: $y = 2 x - 6$ becomes $2 x - y - 6 = 0$ , with $a = 2$ , $b = - 1$ , $c = - 6$ .

Important

Gradient and intercept from the general equation. If $b \ne 0$ , make $y$ the subject:

by = -ax - c \quad\Longrightarrow\quad y = -\frac{a}{b}\,x - \frac{c}{b}, \qquad\text{so}\qquad m = -\frac{a}{b}, \quad y\text{-intercept} = -\frac{c}{b}.

If $b = 0$ , the equation is $a x + c = 0$ , that is, $x = -\frac{c}{a}$ : a vertical line with no gradient. If $a = 0$ , the line is horizontal and $m = 0$ .

Worked example 1: gradient from two points

Find the gradient of the line through (a) $P (- 2, 5)$ and $Q (4, - 1)$ (b) $R (3, 2)$ and $S (3, 7)$ .

Solution (a) $m = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1$ . The line falls 1 unit for each unit across.

(b) $m = \frac{7 - 2}{3 - 3} = \frac{5}{0}$ , which has no value. Both points have $x = 3$ , so $R S$ is the vertical line $x = 3$ , and its gradient is undefined.

Worked example 2: from the general equation to the graph

For the line $3 x - 4 y + 12 = 0$ , find the gradient and both intercepts, and draw it.

Solution Here $a = 3$ , $b = - 4$ , $c = 12$ , so $m = -\frac{3}{-4} = \frac{3}{4}$ .

$y$ -intercept: put $x = 0$ : $- 4 y + 12 = 0$ , so $y = 3$ . $x$ -intercept: put $y = 0$ : $3 x + 12 = 0$ , so $x = - 4$ .

Check a third point: $x = 4$ gives $12 - 4 y + 12 = 0$ , so $y = 6$ ; and from $(0, 3)$ , a run of 4 with a rise of 3 also reaches $(4, 6)$ ✓.

The line through its two intercepts, checked with a third point.

Worked example 3: gradient as a rate of change

A school's water tank holds 2,000 litres. When the tap at the bottom is left open, the volume after $t$ minutes is $V = 2000 - 50 t$ (illustrative figures).

(a) Give the gradient and the $V$ -intercept, and say what each means. (b) When is the tank empty?

Solution (a) The gradient is $- 50$ : the volume falls by 50 litres every minute. The $V$ -intercept is 2,000: the volume at $t = 0$ , a full tank.

(b) Put $V = 0$ : $2000 - 50 t = 0$ , so $t = \frac{2000}{50} = 40$ . The tank is empty after 40 minutes; this is the $t$ -intercept.

Worked example 4: a road gradient in per cent

A road sign gives a gradient as a percentage: 10% means the road rises (or falls) 10 m for every 100 m measured horizontally, a gradient of $\frac{10}{100} = 0.1$ .

(a) A road rises 45 m over a horizontal distance of 600 m. Give its gradient as a decimal and as a percentage. (b) How much height does a lorry lose over 250 m, measured horizontally, of a 10% descent?

Solution (a) $m = \frac{45}{600} = 0.075 = 7.5\%$ . (b) $\text{fall} = 0.1 \times 250 = 25$ m.

Warning

Keep the same order top and bottom. If you start with $y_{2}$ on top, start with $x_{2}$ underneath: $\frac{y_2 - y_1}{x_2 - x_1}$ . Mixing the order, $\frac{y_2 - y_1}{x_1 - x_2}$ , gives the wrong sign.

Rise over run, not run over rise. The change in $y$ goes on top.

Mind the minus sign in $m = -\frac{a}{b}$ . For $2 x + 5 y - 10 = 0$ , $m = -\frac{2}{5}$ , not $\frac{2}{5}$ . If in doubt, make $y$ the subject.

Tip

Intercepts are the quickest way to draw a line from its general form: cover the $y$ term to find the $x$ -intercept, and cover the $x$ term to find the $y$ -intercept. Then check with a third point.

Khan Academy: Slope from equation. Watch how the gradient is read from equations written in several different forms, including by making y the subject.

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Practice

Show your working. The last three are harder.

Find the gradient of the line through $(2, 3)$ and $(6, 11)$ .
Find the gradient of the line through $(- 1, 4)$ and $(3, - 2)$ .
Find the gradient and the $y$ -intercept of $2 x + 5 y - 10 = 0$ .
Write $y = - 3 x + 7$ in the general form $a x + b y + c = 0$ .
Find both intercepts and the gradient of $4 x - 3 y = 12$ , and draw the line.
Say whether each line is horizontal or vertical, and give its equation in simplest form: (a) $2 y - 6 = 0$ (b) $5 x + 10 = 0$ .
A road rises 18 m over a horizontal distance of 150 m. Give its gradient as a percentage.
The points $(1, 2)$ , $(3, 6)$ and $(5, k)$ lie on one straight line. Find $k$ .

Show answersHide answers

$m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2$ .
$m = \frac{-2 - 4}{3 - (-1)} = \frac{-6}{4} = -\frac{3}{2}$ .
$5 y = - 2 x + 10$ , so $y = -\frac{2}{5}x + 2$ : $m = -\frac{2}{5}$ , $y$ -intercept 2.
$3 x + y - 7 = 0$ .
$x$ -intercept $(3, 0)$ , $y$ -intercept $(0, - 4)$ , $m = \frac{4}{3}$ .
(a) horizontal, $y = 3$ (b) vertical, $x = - 2$ .
$\frac{18}{150} = 0.12 = 12\%$ .
The gradient from $(1, 2)$ to $(3, 6)$ is 2. From $(3, 6)$ to $(5, k)$ : $\frac{k - 6}{2} = 2$ , so $k = 10$ .

Sources and credits

Written by Ganzaa.
Photo: Pushing up the hill.jpg, Julius Dehne, Wikimedia Commons, licensed under CC BY-SA 4.0. Used unchanged.
Photo: Warning sign - 10^ gradient ahead, Prendergast, Haverfordwest, Jaggery, via geograph.org.uk and Wikimedia Commons, licensed under CC BY-SA 2.0. Used unchanged.
Video: Slope from equation, Khan Academy, embedded from YouTube.

The graphs, worked examples, practice items and quiz questions were written by Ganzaa; the water-tank figures are illustrative. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1How steep is the road?

How steep is the road?