Coordinates and graphs

Q: What are the key conventions of a 2D Cartesian coordinate grid?

A Cartesian grid has a horizontal $x$-axis and a vertical $y$-axis intersecting at the origin $(0, 0)$, with points written as ordered pairs $(x, y)$.

Q: How do you calculate the midpoint of a line segment joining two coordinates?

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

Q: What formula gives the length or distance between two coordinates on a grid?

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

Q: What do $m$ and $c$ represent in the straight-line equation $y = mx + c$?

$m$ represents the gradient (slope) and $c$ represents the $y$-intercept $(0, c)$.

Q: How is the gradient of a straight line calculated between two points?

$m = \frac{\text{Change in } y}{\text{Change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$

Q: What are the equations and gradients of horizontal and vertical lines?

Horizontal lines have the equation $y = k$ with gradient $m = 0$; vertical lines have the equation $x = k$ with an undefined gradient.

Q: What do the gradient and flat sections represent on a distance–time graph?

The gradient represents speed , and a flat horizontal line indicates the object is stationary (speed = 0).

Q: What do the gradient and area under the curve represent on a velocity–time graph?

The gradient represents acceleration , and the area under the graph represents the total distance travelled .

Q: What are the characteristic shapes of quadratic, cubic, reciprocal, and exponential graphs?

Quadratics form parabolas (U/hill-shapes), cubics form S-shaped curves, reciprocals form two separate asymptotic branches, and exponentials curve steeply upward/downward.

Mathematics 2210 · CCEA GCSE · 8 min read · Updated: 31 Aug 2026

Welcome to Coordinates and Graphs!

Graphs are one of the most powerful tools in Mathematics. They allow us to turn numbers and equations into clear visual pictures. Whether you are tracking the journey of a car, calculating the cost of a mobile phone contract, or finding where two curves meet, graphs bring algebra to life.

Don't worry if you find some parts of algebra tricky. In this guide, we will break down every single topic step by step—from basic plotting on a grid all the way to advanced tangents and circle equations. Let's master this together!

1. Cartesian Coordinates and Line Segments (Units M1, M2, M3)

The Coordinate Grid

A Cartesian coordinate grid is made of two axes crossing at the point $(0, 0)$, which is called the origin:

• The horizontal axis is the $x$-axis (values go left and right).
• The vertical axis is the $y$-axis (values go down and up).

Every point is written as an ordered pair: $(x, y)$.

Memory Trick: "Along the corridor, then up the stairs!" Always read or plot the horizontal $x$-value first, followed by the vertical $y$-value.

Midpoint of a Line Segment

The midpoint is the exact middle point between two coordinates $(x_1, y_1)$ and $(x_2, y_2)$. To find it, simply calculate the average of the $x$-values and the average of the $y$-values:

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

Worked Example:
Find the midpoint between $(-2, 5)$ and $(6, 1)$.
Step 1: Average the $x$-coordinates: $\frac{-2 + 6}{2} = \frac{4}{2} = 2$
Step 2: Average the $y$-coordinates: $\frac{5 + 1}{2} = \frac{6}{2} = 3$
Answer: The midpoint is $(2, 3)$.

Length of a Line Segment

To find the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$, we use Pythagoras' Theorem on a coordinate grid:

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

Worked Example:
Find the length of the line segment joining $(1, 2)$ and $(4, 6)$.
Step 1: Horizontal change $\Delta x = 4 - 1 = 3$
Step 2: Vertical change $\Delta y = 6 - 2 = 4$
Step 3: Apply Pythagoras: $d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
Answer: The length is $5$ units.

Key Takeaway: Coordinates always follow $(x, y)$. The midpoint is the average of coordinates; the distance is found using Pythagoras' theorem on the horizontal and vertical differences.

2. Linear Graphs and Gradients (Units M2, M3, M4)

The Equation of a Straight Line

Every straight-line graph can be written in the standard form:

$$y = mx + c$$

• $m$ is the gradient (the slope or steepness of the line).
• $c$ is the $y$-intercept (the point $(0, c)$ where the line crosses the $y$-axis).

Calculating the Gradient

The gradient measures how much the line rises or falls for each step to the right:

$$m = \frac{\text{Change in } y}{\text{Change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$$

• If the line slopes upwards from left to right, the gradient is positive.
• If the line slopes downwards from left to right, the gradient is negative.

Special Lines to Remember

• Horizontal lines: Have the form $y = k$. Their gradient is $m = 0$. (Example: $y = 4$)
• Vertical lines: Have the form $x = k$. Their gradient is undefined. (Example: $x = -3$)

Parallel and Perpendicular Lines (Unit M4)

Parallel Lines: Lines that run in the same direction never meet. They have the same gradient:

$$m_1 = m_2$$

Example: The line $y = 3x + 1$ is parallel to $y = 3x - 7$.

Perpendicular Lines: Lines that meet at right angles ($90^\circ$). Their gradients are negative reciprocals of each other:

$$m_1 \times m_2 = -1 \quad \implies \quad m_2 = -\frac{1}{m_1}$$

Worked Example:
Find the equation of the line perpendicular to $y = 2x + 5$ that passes through $(0, 4)$.
Step 1: The gradient of the given line is $m_1 = 2$.
Step 2: The perpendicular gradient is $m_2 = -\frac{1}{2}$.
Step 3: The $y$-intercept is given by the point $(0, 4)$, so $c = 4$.
Answer: $y = -\frac{1}{2}x + 4$

Key Takeaway: In $y = mx + c$, $m$ tells you the slope and $c$ tells you the starting height on the vertical axis. Parallel lines have equal gradients ($m_1 = m_2$); perpendicular lines have negative reciprocal gradients ($m_1 \times m_2 = -1$).

3. Real-Life Graphs and Kinematics (Units M2, M6, M7)

Conversion and Cost Graphs

Many real-life situations are represented by straight lines:

• Fixed Charge / Base Value: Represented by the $y$-intercept $c$.
• Rate per Unit: Represented by the gradient $m$.

Example: A plumber charges a £$40$ call-out fee plus £$20$ per hour worked. The cost $C$ for $h$ hours is given by the formula $C = 20h + 40$.

Distance–Time Graphs

A distance–time graph displays how far an object travels over time:

• Gradient = Speed: $\text{Speed} = \frac{\text{Change in Distance}}{\text{Change in Time}}$
• Flat (Horizontal) Line: The object is stationary (not moving, speed $= 0$).
• Steeper Line: Higher speed.
• Line Sloping Downwards: The object is returning towards its starting point.

Velocity–Time Graphs (Higher Tier)

A velocity–time graph shows how an object's speed in a specific direction changes over time:

• Gradient = Acceleration: $\text{Acceleration} = \frac{\text{Change in Velocity}}{\text{Time Taken}}$
• Horizontal Line: Moving at a constant velocity (acceleration $= 0$).
• Area Under the Graph = Total Distance Travelled: Divide the area under the graph into simple geometric shapes (triangles, rectangles, and trapezia) to calculate the distance.

Key Takeaway: On distance–time graphs, gradient = speed. On velocity–time graphs, gradient = acceleration, and the area under the graph = total distance.

4. Non-Linear Graphs and Graphical Solutions (Units M6, M7, M8)

Types of Non-Linear Graphs

1. Quadratic Graphs: $y = ax^2 + bx + c$
These form a symmetrical U-shape (when $a > 0$) or an inverted U-shape / hill shape (when $a < 0$) called a parabola. They have a turning point (vertex) and a vertical line of symmetry.

2. Cubic Graphs: $y = ax^3 + bx^2 + cx + d$
These have an S-shaped curve that can have up to two turning points.

3. Reciprocal Graphs: $y = \frac{k}{x}$ (where $x \neq 0$)
These form two separate curved branches in opposite quadrants. The curve gets closer and closer to the axes but never touches them (the axes are asymptotes).

4. Exponential Graphs: $y = ka^x$ or $y = k^x$ (Unit M8)
These model rapid growth (or decay). The curve rises steeply as $x$ increases and flattens towards the horizontal axis.

Solving Equations Graphically

• Finding Roots: To solve an equation like $ax^2 + bx + c = 0$ graphically, look for the points where the curve crosses the $x$-axis (where $y = 0$).
• Simultaneous Equations: The solution(s) to a pair of equations (for example, a line and a quadratic curve) are the coordinates $(x, y)$ of the points of intersection where the two graphs cross each other.

Important Graphing Rule: When plotting non-linear curves, always join the plotted points with a smooth, continuous, freehand curve. Never use a ruler to connect points on a curve with straight segments!

Key Takeaway: Recognise the characteristic shapes of quadratic (parabolas), cubic (S-curves), reciprocal (split branches), and exponential graphs. Intersections between graphs represent simultaneous solutions.

5. Circles, Tangents, and Rates of Change (Unit M8)

Equation of a Circle

The equation of a circle centred at the origin $(0, 0)$ with radius $r$ is:

$$x^2 + y^2 = r^2$$

Example: A circle with equation $x^2 + y^2 = 49$ has its centre at $(0, 0)$ and a radius of $r = \sqrt{49} = 7$.

Equation of a Tangent to a Circle

A tangent is a straight line that touches a circle at exactly one point. A fundamental geometric rule is that the tangent is perpendicular to the radius at that point of contact.

Step-by-Step Method to Find the Tangent Equation at $(x_1, y_1)$:
1. Find the gradient of the radius from $(0, 0)$ to $(x_1, y_1)$:
$$m_{\text{radius}} = \frac{y_1 - 0}{x_1 - 0} = \frac{y_1}{x_1}$$
2. Find the gradient of the tangent (negative reciprocal):
$$m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{x_1}{y_1}$$
3. Use the point-slope formula $y - y_1 = m_{\text{tangent}}(x - x_1)$ or substitute into $y = mx + c$ to find $c$.

Worked Example:
Find the equation of the tangent to the circle $x^2 + y^2 = 25$ at the point $(3, 4)$.
Step 1: Gradient of radius $= \frac{4}{3}$.
Step 2: Gradient of tangent $= -\frac{3}{4}$.
Step 3: Substitute $(3, 4)$ into $y = mx + c$:
$4 = \left(-\frac{3}{4}\right)(3) + c$
$4 = -\frac{9}{4} + c \implies c = 4 + 2.25 = 6.25 = \frac{25}{4}$
Answer: $y = -\frac{3}{4}x + \frac{25}{4}$ (or $3x + 4y = 25$).

Instantaneous Rate of Change

To find the rate of change (gradient) at a specific point on a non-linear curve:
1. Place a ruler at the specific point and draw a single, balanced tangent line that touches the curve without cutting through it.
2. Pick two convenient points on your tangent line.
3. Calculate the gradient using $m = \frac{\Delta y}{\Delta x}$.

Key Takeaway: For a circle $x^2 + y^2 = r^2$, the tangent at any point is perpendicular to the radius connecting that point to the origin. To find the gradient of any curve at a point, draw a tangent and calculate its slope.

6. Common Pitfalls & Examiner Tips

• Mixing up $x$ and $y$: Remember that $(0, 3)$ is on the $y$-axis, while $(3, 0)$ is on the $x$-axis. Double-check your coordinates!

• Counting Squares Instead of Reading Scales: When finding gradients on exam papers, always use the numbers marked on the axes, not just the number of grid squares. The axes may have different scales!

• Forgetting Negative Slopes: If a line goes downwards from left to right, its gradient must be negative.

• Perpendicular Gradient Errors: Remember both steps: flip the fraction and change the sign! (e.g., $4 \implies -\frac{1}{4}$, $-\frac{2}{3} \implies \frac{3}{2}$).

• Rough Curves: Use a sharp pencil and draw smooth curves in one continuous stroke. Do not join points with straight ruler lines unless the graph is explicitly linear.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What are the key conventions of a 2D Cartesian coordinate grid?

A Cartesian grid has a horizontal $x$-axis and a vertical $y$-axis intersecting at the origin $(0, 0)$, with points written as ordered pairs $(x, y)$.

How do you calculate the midpoint of a line segment joining two coordinates?

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

What formula gives the length or distance between two coordinates on a grid?

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

What do $m$ and $c$ represent in the straight-line equation $y = mx + c$?

$m$ represents the gradient (slope) and $c$ represents the $y$-intercept $(0, c)$.

How is the gradient of a straight line calculated between two points?

$m = \frac{\text{Change in } y}{\text{Change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$

What are the equations and gradients of horizontal and vertical lines?

Horizontal lines have the equation $y = k$ with gradient $m = 0$; vertical lines have the equation $x = k$ with an undefined gradient.

What is the relationship between the gradients of parallel lines and perpendicular lines?

Parallel lines have equal gradients ($m_1 = m_2$); perpendicular lines have negative reciprocal gradients ($m_1 \times m_2 = -1$ or $m_2 = -\frac{1}{m_1}$).

What do the gradient and flat sections represent on a distance–time graph?

The gradient represents speed, and a flat horizontal line indicates the object is stationary (speed = 0).

What do the gradient and area under the curve represent on a velocity–time graph?

The gradient represents acceleration, and the area under the graph represents the total distance travelled.

What are the characteristic shapes of quadratic, cubic, reciprocal, and exponential graphs?

Quadratics form parabolas (U/hill-shapes), cubics form S-shaped curves, reciprocals form two separate asymptotic branches, and exponentials curve steeply upward/downward.

How can equations and simultaneous equations be solved using graphs?

Roots of $f(x) = 0$ are the $x$-intercepts; simultaneous solutions are the $(x, y)$ coordinates where the graphs intersect.

What is the standard equation of a circle centred at the origin?

$x^2 + y^2 = r^2$, where $r$ is the radius of the circle.

How do you find the equation of a tangent to a circle at a given point $(x_1, y_1)$?

Find the radius gradient $m_{\text{radius}} = \frac{y_1}{x_1}$, take its negative reciprocal $m_{\text{tangent}} = -\frac{x_1}{y_1}$, and substitute $(x_1, y_1)$ into $y = mx + c$.

How do you estimate the instantaneous rate of change at a point on a non-linear curve?

Draw a straight tangent line touching the curve at that point, then calculate the gradient of that tangent line using $m = \frac{\Delta y}{\Delta x}$.

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