What's the difference between $y=mx+c$ and $y-y_1=m(x-x_1)$?

$y=mx+c$ is useful when you know the gradient and the y-intercept. $y-y_1=m(x-x_1)$ is more general and powerful; you can use it whenever you know the gradient and *any* point on the line, not just the y-intercept. Most problems are solved more quickly starting with the point-gradient form.

My answer for a line equation was $y = 0.5x + 2.5$ but the mark scheme said $x - 2y + 5 = 0$. Is my answer wrong?

Not necessarily. Your answer is mathematically correct, but exam questions often specify the format of the final answer, typically $ax+by+c=0$ where $a, b, c$ are integers. To convert your answer, multiply everything by 2 to remove the decimals: $2y = x + 5$. Then rearrange to get everything on one side: $x - 2y + 5 = 0$. Always check the required format.

How do I know whether to use the distance formula or the gradient formula to prove a property of a shape?

Think about the geometric property you need to prove. If it involves lengths (e.g., 'show the triangle is isosceles', 'show it is a rhombus'), use the distance formula. If it involves angles, especially right angles (e.g., 'show it is a right-angled triangle', 'show it is a rectangle'), use the gradient formula to check for perpendicular lines ($m_1m_2 = -1$).

What does an 'undefined' gradient mean?

An undefined gradient occurs when you try to calculate the gradient of a vertical line. For points $(x_1, y_1)$ and $(x_1, y_2)$, the formula becomes $\frac{y_2-y_1}{x_1-x_1} = \frac{y_2-y_1}{0}$. Since division by zero is undefined, so is the gradient. The equation of such a line is simply $x = k$, where $k$ is the constant x-coordinate.

Where can I learn Coordinate geometry for Cambridge 9709?

This free lesson on MarkScheme covers syllabus point 1.3 (Coordinate geometry) for Cambridge A-Level Mathematics. It includes visual revision, key points, and links to real past-paper questions with mark-scheme marking at markscheme.app/courses/9709/1-3-coordinate-geometry.

Is Coordinate geometry (1.3) on 9709 free to study?

Yes. The Mathematics 9709 course on MarkScheme is 100% free — including premium visual lessons, exam tips, and past-paper marking. No paywall for Coordinate geometry.

Coordinate geometry 9709 — Free A-Level Maths

Coordinate geometry

Coordinate geometry is the bridge between algebra and geometry. It allows us to use algebraic equations to describe and analyse geometric figures like lines, circles, and polygons on a Cartesian plane. This topic is fundamental, providing the tools to solve a vast range of problems in pure mathematics and its applications.

BY THE END, YOU CAN…

1Calculate the gradient, length, and midpoint of a line segment given the coordinates of its endpoints.
2Find the equation of a straight line in the form y = mx + c or ax + by + c = 0.
3Understand and use the relationship between the gradients of parallel and perpendicular lines.
4Apply coordinate geometry methods to solve problems involving points, lines, and shapes on a Cartesian plane.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Geometry of Grids

Coordinate geometry uses algebra to describe geometric shapes. By placing shapes on a grid, we can use equations and formulas to analyse their properties like length, slope, and position.

THINK OF IT LIKE…

Imagine you're giving directions in a city built on a perfect grid. The 'gradient' is how steep a street is. 'Parallel' streets run side-by-side and never meet, so they have the same steepness. The 'distance' is the actual length you walk between two points, and the 'midpoint' is the halfway house on your journey.

1
A straight line is defined by y = mx + c, where 'm' is the gradient (steepness) and 'c' is the y-intercept (where it crosses the vertical axis).
2
Parallel lines share the same gradient (m₁ = m₂). Perpendicular lines meet at a 90° angle, and their gradients multiply to give -1 (m₁m₂ = -1).
3
The distance between two points (x₁, y₁) and (x₂, y₂) is found using Pythagoras' theorem: √((x₂−x₁)² + (y₂−y₁)²).
4
The midpoint of the line segment connecting two points is the average of their coordinates: ((x₁+x₂)/2, (y₁+y₂)/2).

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

INTERACTIVESTEP 1 / 3 · The line

Drag A and B — a unique straight line passes through the two points.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

y = mx +

Tap a symbol — great for exam definitions

$m_1$

$(x_1, y_1)$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

The Equation of a Straight Line

Every straight line on a 2D plane can be represented by an equation. The two most useful forms you will encounter are the gradient-intercept form and the point-gradient form. Understanding how to switch between them and when to use each is a key skill.

Gradient-Intercept Form: $y = mx + c$

Where $m$ is the gradient and $c$ is the y-intercept.

Point-Gradient Form: $y - y_1 = m(x - x_1)$

Where $m$ is the gradient and $(x_1, y_1)$ is any point on the line.

The gradient $m$ measures the steepness of the line. It is calculated as 'rise over run': $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
A positive gradient means the line slopes upwards from left to right.
A negative gradient means the line slopes downwards from left to right.
A horizontal line has a gradient of 0. A vertical line has an undefined gradient.
The final equation is often required in the form $ax + by + c = 0$ , where $a, b, c$ are integers.

Parallel and Perpendicular Lines

The relationship between lines can be determined entirely by their gradients. Parallel lines travel in the same direction and thus have the same steepness. Perpendicular lines intersect at a right angle (90°), and their gradients have a special relationship.

For two lines with gradients $m_1$ and $m_2$ :

Parallel condition: $m_1 = m_2$

Perpendicular condition: $m_1 \times m_2 = -1$

EXAM TIP

For perpendicular lines, remember the phrase 'negative reciprocal'. To find the perpendicular gradient, you flip the fraction and change the sign. For example, if $m_1 = \frac{2}{5}$ , the perpendicular gradient $m_2$ is $-\frac{5}{2}$ . If $m_1 = -3$ , then $m_2 = -\frac{1}{-3} = +\frac{1}{3}$ .

Distance, Midpoint and Geometric Applications

We can also calculate key properties of line segments. The distance formula, derived from Pythagoras' theorem, gives the length of a segment. The midpoint formula finds the coordinates of the point exactly halfway between two endpoints. These tools are essential for proving properties of geometric shapes, such as showing a triangle is isosceles or that the diagonals of a square bisect each other.

Distance between $(x_1, y_1)$ and $(x_2, y_2)$ : $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Midpoint of segment between $(x_1, y_1)$ and $(x_2, y_2)$ : $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

The line $L_1$ passes through the points $A(2, 5)$ and $B(8, 2)$ . A second line, $L_2$ , is perpendicular to $L_1$ and passes through point $A$ . Find the equation of $L_2$ , giving your answer in the form $ax+by+c=0$ where $a, b, c$ are integers.

1
Step 1: Find the gradient of line $L_1$ . Gradient of $L_1$ , $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 5}{8 - 2} = \frac{-3}{6} = -\frac{1}{2}$ . [M1 for gradient formula]

EXAMPLE 2

The points $P$ , $Q$ , and $R$ have coordinates $(-1, 6)$ , $(3, 0)$ , and $(9, 4)$ respectively. Find the equation of the perpendicular bisector of the line segment $PQ$ .

1
Step 1: Find the midpoint of $PQ$ . Midpoint $M = \left( \frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2} \right) = \left( \frac{-1 + 3}{2}, \frac{6 + 0}{2} \right) = \left( \frac{2}{2}, \frac{6}{2} \right) = (1, 3)$ . [M1 for midpoint formula]

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEACoordinate geometry

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 12

What is the formula for the gradient, $m$ , of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ ?

TAP OR SPACE TO FLIP

ANSWER

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

TAP OR SPACE TO FLIP BACK

1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
The gradient $m$ measures the steepness of the line. It is calculated as 'rise over run': $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
✓
A positive gradient means the line slopes upwards from left to right.
✓
A negative gradient means the line slopes downwards from left to right.
✓
A horizontal line has a gradient of 0. A vertical line has an undefined gradient.
✓
The final equation is often required in the form $ax + by + c = 0$ , where $a, b, c$ are integers.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Practice Coordinate Geometry Problems

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Practice Coordinate Geometry Problems on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS1.2 Functions NEXT →1.4 Circular measure

Coordinate geometry

BY THE END, YOU CAN…

1Calculate the gradient, length, and midpoint of a line segment given the coordinates of its endpoints.
2Find the equation of a straight line in the form y = mx + c or ax + by + c = 0.
3Understand and use the relationship between the gradients of parallel and perpendicular lines.
4Apply coordinate geometry methods to solve problems involving points, lines, and shapes on a Cartesian plane.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Geometry of Grids

Coordinate geometry uses algebra to describe geometric shapes. By placing shapes on a grid, we can use equations and formulas to analyse their properties like length, slope, and position.

THINK OF IT LIKE…

1
A straight line is defined by y = mx + c, where 'm' is the gradient (steepness) and 'c' is the y-intercept (where it crosses the vertical axis).
2
Parallel lines share the same gradient (m₁ = m₂). Perpendicular lines meet at a 90° angle, and their gradients multiply to give -1 (m₁m₂ = -1).
3
The distance between two points (x₁, y₁) and (x₂, y₂) is found using Pythagoras' theorem: √((x₂−x₁)² + (y₂−y₁)²).
4
The midpoint of the line segment connecting two points is the average of their coordinates: ((x₁+x₂)/2, (y₁+y₂)/2).

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

INTERACTIVESTEP 1 / 3 · The line

Drag A and B — a unique straight line passes through the two points.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

y = mx +

Tap a symbol — great for exam definitions

$m_1$

$(x_1, y_1)$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

The Equation of a Straight Line

Gradient-Intercept Form: $y = mx + c$

Where $m$ is the gradient and $c$ is the y-intercept.

Point-Gradient Form: $y - y_1 = m(x - x_1)$

Where $m$ is the gradient and $(x_1, y_1)$ is any point on the line.

The gradient $m$ measures the steepness of the line. It is calculated as 'rise over run': $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
A positive gradient means the line slopes upwards from left to right.
A negative gradient means the line slopes downwards from left to right.
A horizontal line has a gradient of 0. A vertical line has an undefined gradient.
The final equation is often required in the form $ax + by + c = 0$ , where $a, b, c$ are integers.

Parallel and Perpendicular Lines

For two lines with gradients $m_1$ and $m_2$ :

Parallel condition: $m_1 = m_2$

Perpendicular condition: $m_1 \times m_2 = -1$

EXAM TIP

Distance, Midpoint and Geometric Applications

Distance between $(x_1, y_1)$ and $(x_2, y_2)$ : $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Midpoint of segment between $(x_1, y_1)$ and $(x_2, y_2)$ : $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

1
Step 1: Find the gradient of line $L_1$ . Gradient of $L_1$ , $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 5}{8 - 2} = \frac{-3}{6} = -\frac{1}{2}$ . [M1 for gradient formula]

EXAMPLE 2

The points $P$ , $Q$ , and $R$ have coordinates $(-1, 6)$ , $(3, 0)$ , and $(9, 4)$ respectively. Find the equation of the perpendicular bisector of the line segment $PQ$ .

1
Step 1: Find the midpoint of $PQ$ . Midpoint $M = \left( \frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2} \right) = \left( \frac{-1 + 3}{2}, \frac{6 + 0}{2} \right) = \left( \frac{2}{2}, \frac{6}{2} \right) = (1, 3)$ . [M1 for midpoint formula]

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEACoordinate geometry

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 12

What is the formula for the gradient, $m$ , of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ ?

TAP OR SPACE TO FLIP

ANSWER

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

TAP OR SPACE TO FLIP BACK

1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
The gradient $m$ measures the steepness of the line. It is calculated as 'rise over run': $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
✓
A positive gradient means the line slopes upwards from left to right.
✓
A negative gradient means the line slopes downwards from left to right.
✓
A horizontal line has a gradient of 0. A vertical line has an undefined gradient.
✓
The final equation is often required in the form $ax + by + c = 0$ , where $a, b, c$ are integers.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Practice Coordinate Geometry Problems

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Practice Coordinate Geometry Problems on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS1.2 Functions NEXT →1.4 Circular measure

Revise Coordinate geometry for Cambridge 9709 (1.3)

In simple terms

Explore the concept

Key formulas

Full topic notes

The Equation of a Straight Line

Parallel and Perpendicular Lines

Distance, Midpoint and Geometric Applications

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Practice Coordinate Geometry Problems

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

The Equation of a Straight Line

Parallel and Perpendicular Lines

Distance, Midpoint and Geometric Applications

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Practice Coordinate Geometry Problems

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Coordinate geometry