3.1 Straight lines, gradients and straight-line models — Learn

Prerequisites: plotting coordinates; reading a gradient as rise over run; rearranging a formula; solving a linear equation.

By the end of this section you can: work out the gradient of a line from two points; write the equation of a line from a gradient and a point, and from two points; move between the forms $y = mx + c$, $y - y_1 = m(x - x_1)$ and $ax + by + c = 0$; read the gradient and the intercept of a straight-line model with their units.

1. The gradient of a line

The gradient of a line measures how steeply it rises. It is the change in $y$ for each unit of change in $x$. You can read it as rise over run.

For two points $(x_1, y_1)$ and $(x_2, y_2)$ on a line,

The order matters. Subtract the first coordinates from the second, on the top and on the bottom, or the sign comes out wrong.

A line that rises from left to right has a positive gradient. A line that falls from left to right has a negative gradient. A horizontal line has gradient $0$. A vertical line has no gradient: dividing by a zero change in $x$ is not allowed.

Example 1. Work out the gradient of the line through $(2, 5)$ and $(7, 11)$.

Subtract the coordinates of the first point from the second:

The line rises $6$ units for every $5$ units across, so the gradient is $\frac{6}{5}$.

Example 2. Work out the gradient of the line through $(-3, 4)$ and $(5, -2)$.

The gradient is negative, so the line falls as it moves right.

2. The equation of a line

Two facts fix a straight line: one point on it, and its gradient. The equation of the line through $(x_1, y_1)$ with gradient $m$ is

You can also write every non-vertical line in the form

Here $m$ is the gradient and $c$ is the intercept on the $y$-axis, the value of $y$ where the line crosses it.

Example 3. Find the equation of the line with gradient $4$ through $(3, -2)$.

Substitute the point and the gradient into $y - y_1 = m(x - x_1)$:

The subtraction of a negative gives a plus on the left. Expand the bracket:

Subtract $2$ from both sides:

The line cuts the $y$-axis at $(0, -14)$.

Example 4. Find the equation of the line through $(2, 5)$ and $(7, 11)$.

The gradient is $\frac{6}{5}$, from Example 1. Substitute one point, here $(2, 5)$:

Expand the bracket:

Add $5$, which is $\frac{25}{5}$, to both sides:

So $c = \frac{13}{5}$. Check the answer by substituting the other point, $(7, 11)$: $\frac{6}{5} \times 7 + \frac{13}{5} = \frac{42 + 13}{5} = 11$. Both points fit, so the equation is right.

3. The form $ax + by + c = 0$

A line can also be written with all the terms on one side:

This form handles vertical lines as well, and marks often ask for it with whole numbers. Clear the fractions by multiplying every term.

Take Example 4 and multiply by $5$:

Collect everything on the left:

To find where a line crosses the axes, substitute $y = 0$ for the $x$-intercept and $x = 0$ for the $y$-intercept.

Example 5. Find where $2x + 3y = 12$ crosses the axes.

Put $y = 0$:

Put $x = 0$:

The line crosses the $x$-axis at $(6, 0)$ and the $y$-axis at $(0, 4)$.

4. Straight-line models

Many real quantities are linked by a straight line. The equation then carries meaning:

• the gradient is the rate of change, in units of the vertical quantity per unit of the horizontal quantity;
• the intercept is the starting value, when the horizontal quantity is $0$.

Example 6. A taxi charges a fixed £3 plus £2 for each kilometre. Write a model for the cost, and say what the gradient and the intercept mean.

Let $d$ be the distance in kilometres and $C$ the cost in pounds:

The gradient is $2$. It costs £2 for each extra kilometre. The intercept is $3$. It is the fixed charge, paid before the taxi moves. For a journey of $8$ km,

The fare is £19.

The units of a gradient are the units of $y$ divided by the units of $x$. Here the gradient is $2$ pounds per kilometre.

5. Try it: the line explorer

Predict first. You will change the gradient $m$ and the intercept $c$ in $y = mx + c$. Before you touch the controls, predict what happens to the line when $m$ is negative, and what happens when $c$ changes but $m$ stays the same. Write both predictions down.

Test. Move the slider for $m$ and watch the direction of the line. Then keep $m$ fixed and move the slider for $c$. The readout names the gradient, the intercept, and the equation in the form $ax + by + c = 0$.

Explain. For one setting, read the step drawn on the graph: across $1$ and up by the gradient. Write two sentences saying why changing $c$ moves the whole line up or down without tilting it.

Watch out. Three slips are common on this page.

The sign of the gradient. A line that falls as it moves right has a negative gradient. Read the direction before you calculate, and expect a minus sign.

The sign inside the bracket. The form $y - y_1 = m(x - x_1)$ subtracts both coordinates. For the point $(2, 5)$ the left side is $y - 5$, not $y + 5$. A quick substitution of the point back into your answer catches this at once.

The two intercepts. The $y$-intercept is $c$ in $y = mx + c$, at the point $(0, c)$. The $x$-intercept is where $y = 0$, and it is a different number. Re-check: after finding an equation, substitute both given points, and confirm that each one makes the two sides equal.

Progress check

Return to the chapter problem. The centre line of the causeway runs through $(12, 7)$ and $(5, 14)$.

1. Work out the gradient of the causeway.
2. Write the equation of the causeway in the form $y = mx + c$.
3. Rewrite it in the form $x + y = k$.
4. The lake's rim passes through $(11, 8)$ and $(6, 13)$. Substitute each point into your equation to check that both lie on the causeway.

You should now be able to

• work out the gradient of a line from two points, with the correct sign;
• find the equation of a line from a gradient and a point, and from two points;
• move between $y = mx + c$, $y - y_1 = m(x - x_1)$ and $ax + by + c = 0$;
• find where a line crosses each axis;
• read the gradient and the intercept of a straight-line model and give their units;
• check a line equation by substituting a point back into it.