Linear graphs

What you will learn

sketch linear graphs from gradient-intercept form $y = mx + c$ ,
convert general form $ax + by = c$ to gradient-intercept form and sketch,
calculate the gradient between two points using $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ ,
find the midpoint of a line segment using $M = \left(\dfrac{x_1 + x_2}{2},\;\dfrac{y_1 + y_2}{2}\right)$ ,
calculate the distance between two points using Pythagoras,
identify parallel lines (equal gradients) and perpendicular lines (gradient product $= -1$ ).

Worked example 0 Real-world example: comparing phone plans

Plan A charges a $10 monthly fee plus $0.50 per GB. Plan B charges no monthly fee but $1.20 per GB. Which plan is cheaper at $20$ GB?

Plan A cost: $C_A = 0.5n + 10$ . At $n = 20$ : $C_A = 0.5(20) + 10 = 20$ .
Plan B cost: $C_B = 1.2n$ . At $n = 20$ : $C_B = 1.2(20) = 24$ .
Plan A is $4 cheaper at $20$ GB. Graphing both lines would show them crossing near $n \approx 14.3$ GB — below that, Plan B is cheaper.

Key idea: the gradient tells you the per-GB rate; the $y$ -intercept tells you the fixed cost. Comparing gradients and intercepts lets you decide which plan suits your usage.

1. Gradient-intercept form: $y = mx + c$

In the equation $y = mx + c$ :

$m$ is the gradient (slope) — how steeply the line rises or falls,
$c$ is the $y$ -intercept — where the line crosses the $y$ -axis.

Formula reference

y = mx + c

$m$ = gradient (rise/run), $c$ = $y$ -intercept.

To sketch from $y = mx + c$ :

Plot the $y$ -intercept $(0, c)$ .
From that point, use the gradient: move $1$ unit right and $m$ units up (or down if $m < 0$ ).
Draw a straight line through the two points and extend in both directions.

Worked example 1 Sketching from y = mx + c

Sketch $y = 2x - 3$ .

The $y$ -intercept is $-3$ , so plot $(0, -3)$ .
Gradient $m = 2 = \dfrac{2}{1}$ : from $(0, -3)$ move $1$ right and $2$ up to $(1, -1)$ .
Draw the line through $(0, -3)$ and $(1, -1)$ .
The $x$ -intercept is found by setting $y = 0$ : $0 = 2x - 3$ , so $x = 1.5$ . Confirm the line passes through $(1.5, 0)$ .

2. General form and converting between forms

The general form of a linear equation is $ax + by = c$ (or $ax + by + c = 0$ ). To sketch it, rearrange to gradient-intercept form.

Worked example 2 Converting general form to y = mx + c

Rewrite $3x + 2y = 12$ in gradient-intercept form and state the gradient and $y$ -intercept.

Subtract $3x$ : $2y = -3x + 12$ .
Divide by $2$ : $y = -\dfrac{3}{2}x + 6$ .
Gradient $m = -\dfrac{3}{2}$ , $y$ -intercept $c = 6$ .

Worked example 3 Sketching using x- and y-intercepts

Sketch $4x + 3y = 24$ .

Set $x = 0$ : $3y = 24$ , so $y = 8$ . The $y$ -intercept is $(0, 8)$ .
Set $y = 0$ : $4x = 24$ , so $x = 6$ . The $x$ -intercept is $(6, 0)$ .
Plot both intercepts and draw the straight line.

3. Gradient from two points

Formula reference

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

Worked example 4 Gradient from two points

Find the gradient of the line through $(2, 3)$ and $(8, 15)$ .

m = \frac{15 - 3}{8 - 2} = \frac{12}{6} = 2

The line rises $2$ units for every $1$ unit across.

Worked example 5 Negative gradient

Find the gradient of the line through $(-1, 7)$ and $(3, -5)$ .

m = \frac{-5 - 7}{3 - (-1)} = \frac{-12}{4} = -3

A negative gradient means the line falls from left to right.

4. Midpoint and distance formulas

Midpoint

The midpoint of a segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is:

Formula reference

M = \left(\frac{x_1 + x_2}{2},\;\frac{y_1 + y_2}{2}\right)

Worked example 6 Midpoint between two points

Find the midpoint of $A(1, 4)$ and $B(7, 10)$ .

M = \left(\frac{1 + 7}{2},\;\frac{4 + 10}{2}\right) = (4,\;7)

Distance

The distance between two points uses Pythagoras’ theorem on the horizontal and vertical differences:

Formula reference

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Worked example 7 Distance between two points

Find the distance between $P(2, 1)$ and $Q(6, 4)$ .

d = \sqrt{(6-2)^2 + (4-1)^2} = \sqrt{16 + 9} = \sqrt{25} = 5

5. Parallel and perpendicular lines

Parallel lines have the same gradient: $m_1 = m_2$ .

Perpendicular lines have gradients whose product is $-1$ : $m_1 \times m_2 = -1$ .

Equivalently, if one line has gradient $m$ , a perpendicular line has gradient $-\dfrac{1}{m}$ .

Worked example 8 Testing parallel or perpendicular

Line $A$ : $y = 3x + 1$ . Line $B$ : $y = -\dfrac{1}{3}x + 5$ .

Gradient of $A$ : $m_A = 3$ .
Gradient of $B$ : $m_B = -\dfrac{1}{3}$ .
Product: $3 \times \left(-\dfrac{1}{3}\right) = -1$ .
Since the product is $-1$ , the lines are perpendicular.

Worked example 9 Finding a parallel line through a given point

Find the equation of the line parallel to $y = 4x - 7$ that passes through $(2, 5)$ .

A parallel line has the same gradient: $m = 4$ .
Substitute into $y = mx + c$ : $5 = 4(2) + c$ , so $c = -3$ .
Equation: $y = 4x - 3$ .

Practice

Fluency

Tier 1: gradient, intercept and sketching

State the gradient and $y$ -intercept of $y = 5x - 2$ .
State the gradient and $y$ -intercept of $y = -3x + 7$ .
Rewrite $2x + y = 10$ in the form $y = mx + c$ .
Rewrite $6x - 3y = 9$ in gradient-intercept form.
Find the $x$ - and $y$ -intercepts of $5x + 2y = 20$ .
Find the gradient of the line through $(1, 2)$ and $(4, 11)$ .
Find the gradient of the line through $(-3, 5)$ and $(1, -7)$ .
Find the midpoint of $(2, 6)$ and $(10, 14)$ .
Find the distance between $(0, 0)$ and $(5, 12)$ .
Find the distance between $(-1, 3)$ and $(2, 7)$ .

Reasoning

Tier 2: mixed practice

A line passes through $(0, -4)$ with gradient $\dfrac{2}{3}$ . Write its equation.
Find the equation of the line through $(1, 5)$ and $(3, 11)$ .
Determine whether the lines $y = 2x + 3$ and $4x - 2y = 7$ are parallel, perpendicular, or neither.
Find the midpoint and length of the segment joining $A(-2, 3)$ and $B(6, -1)$ .
A line has equation $3x + 4y = 24$ . Find its gradient, $x$ -intercept and $y$ -intercept.
Find the equation of the line perpendicular to $y = -2x + 1$ that passes through $(4, 3)$ .
Show that the triangle with vertices $A(0, 0)$ , $B(4, 0)$ and $C(4, 3)$ is right-angled by calculating all three side lengths.
The midpoint of $P(a, 3)$ and $Q(5, 9)$ is $(4, 6)$ . Find the value of $a$ .

Reasoning

Tier 3: explain and apply

Explain why a vertical line cannot be written in the form $y = mx + c$ . What happens to the gradient formula when $x_1 = x_2$ ?
Two hikers start at point $A(1, 2)$ on a grid map (km units). Hiker 1 walks to $B(7, 10)$ . Hiker 2 walks to $C(9, 5)$ . Who walks further, and by how much?
A quadrilateral has vertices $P(0, 0)$ , $Q(6, 0)$ , $R(8, 4)$ , $S(2, 4)$ . By calculating gradients, show that $PQRS$ is a parallelogram.
The line $y = kx + 2$ is perpendicular to $y = 4x - 1$ . Find $k$ .
Point $M(3, 5)$ is the midpoint of $A(1, 2)$ and $B$ . Find the coordinates of $B$ .

Challenge

Reasoning

Harder reasoning

Prove that the diagonals of the rectangle with vertices $A(0, 0)$ , $B(8, 0)$ , $C(8, 6)$ , $D(0, 6)$ bisect each other by finding both midpoints.
A line passes through $(2, -1)$ and is perpendicular to the line $3x - 5y = 10$ . Find its equation in general form $ax + by = c$ .
Three points are $A(1, 1)$ , $B(5, 3)$ , $C(3, 5)$ . Find the perimeter of triangle $ABC$ , giving your answer in exact (surd) form.
The vertices of a triangle are $P(-1, 4)$ , $Q(5, 0)$ and $R(3, 8)$ . Find the length of the median from $P$ to the midpoint of $QR$ .

Interactive

Try it yourself: build three lines

Work through 3 examples on one graph. Drag, check, then move to the next.

Example 1 (easy). Make your line pass through the red target at (2, 4). Many m and c work — any will do.

Point A: (0, 0)
Point B: (2, 2)
Slope m: 1
Intercept c: 0

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Year 9 core - answers

Fluency

Tier 1: gradient, intercept and sketching

Gradient $= 5$ , $y$ -intercept $= -2$ .
Gradient $= -3$ , $y$ -intercept $= 7$ .
$y = -2x + 10$ .
$y = 2x - 3$ .
$x$ -intercept: $(4, 0)$ ; $y$ -intercept: $(0, 10)$ .
$m = \dfrac{11 - 2}{4 - 1} = \dfrac{9}{3} = 3$ .
$m = \dfrac{-7 - 5}{1 - (-3)} = \dfrac{-12}{4} = -3$ .
$M = \left(\dfrac{2+10}{2},\;\dfrac{6+14}{2}\right) = (6, 10)$ .
$d = \sqrt{25 + 144} = \sqrt{169} = 13$ .
$d = \sqrt{9 + 16} = \sqrt{25} = 5$ .

Reasoning

Tier 2: mixed practice

$y = \dfrac{2}{3}x - 4$ .
Gradient: $m = \dfrac{11-5}{3-1} = 3$ . Using $(1, 5)$ : $5 = 3(1) + c$ , $c = 2$ . Equation: $y = 3x + 2$ .
Rearrange $4x - 2y = 7$ : $y = 2x - \dfrac{7}{2}$ . Gradient $= 2$ . Same gradient as $y = 2x + 3$ , so the lines are parallel.
Midpoint: $\left(\dfrac{-2+6}{2},\;\dfrac{3+(-1)}{2}\right) = (2, 1)$ . Length: $\sqrt{8^2 + 4^2} = \sqrt{80} = 4\sqrt{5}$ .
Rearrange: $y = -\dfrac{3}{4}x + 6$ . Gradient $= -\dfrac{3}{4}$ . $y$ -intercept: $(0, 6)$ . $x$ -intercept: $(8, 0)$ .
Gradient of perpendicular: $m = \dfrac{1}{2}$ . Using $(4, 3)$ : $3 = \dfrac{1}{2}(4) + c$ , $c = 1$ . Equation: $y = \dfrac{1}{2}x + 1$ .
$AB = 4$ , $BC = 3$ , $AC = \sqrt{16 + 9} = 5$ . Check: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ . Right-angled at $B$ .
Midpoint $x$ -coordinate: $\dfrac{a + 5}{2} = 4$ , so $a + 5 = 8$ , $a = 3$ .

Reasoning

Tier 3: explain and apply

A vertical line has the form $x = k$ . In the gradient formula, $x_1 = x_2$ makes the denominator zero, so $m$ is undefined. Since $y = mx + c$ requires a defined $m$ , vertical lines cannot be written in this form.
Hiker 1: $d = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$ km. Hiker 2: $d = \sqrt{8^2 + 3^2} = \sqrt{73} \approx 8.54$ km. Hiker 1 walks further by $(10 - \sqrt{73}) \approx 1.46$ km.
Gradient of $PQ$ : $\dfrac{0}{6} = 0$ . Gradient of $SR$ : $\dfrac{0}{6} = 0$ . Gradient of $PS$ : $\dfrac{4}{2} = 2$ . Gradient of $QR$ : $\dfrac{4}{2} = 2$ . Opposite sides have equal gradients, so $PQRS$ is a parallelogram.
Perpendicular gradients: $k \times 4 = -1$ , so $k = -\dfrac{1}{4}$ .
Midpoint formula: $3 = \dfrac{1 + x_B}{2}$ gives $x_B = 5$ ; $5 = \dfrac{2 + y_B}{2}$ gives $y_B = 8$ . So $B = (5, 8)$ .

Reasoning

Challenge

Midpoint of $AC$ : $\left(\dfrac{0+8}{2},\;\dfrac{0+6}{2}\right) = (4, 3)$ . Midpoint of $BD$ : $\left(\dfrac{8+0}{2},\;\dfrac{0+6}{2}\right) = (4, 3)$ . Both midpoints are the same, so the diagonals bisect each other.
Gradient of $3x - 5y = 10$ is $\dfrac{3}{5}$ . Perpendicular gradient: $-\dfrac{5}{3}$ . Through $(2, -1)$ : $y + 1 = -\dfrac{5}{3}(x - 2)$ . Multiply by $3$ : $3y + 3 = -5x + 10$ . Rearrange: $5x + 3y = 7$ .
$AB = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$ . $BC = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ . $AC = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$ . Perimeter $= 4\sqrt{5} + 2\sqrt{2}$ .
Midpoint of $QR$ : $\left(\dfrac{5+3}{2},\;\dfrac{0+8}{2}\right) = (4, 4)$ . Distance from $P(-1, 4)$ to $(4, 4)$ : $\sqrt{25 + 0} = 5$ .

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What you will learn

1. Gradient-intercept form: y=mx+cy = mx + cy=mx+c

Formula reference

2. General form and converting between forms

3. Gradient from two points

Formula reference

4. Midpoint and distance formulas

Midpoint

Formula reference

Distance

Formula reference

5. Parallel and perpendicular lines

Practice

Challenge

Try it yourself: build three lines

Answer key

Year 9 core - answers

1. Gradient-intercept form: $y = mx + c$