Home > CSEC Mathematics > Cartesian Coordinate System

Cartesian Coordinate System

CSEC Mathematics: Introduction to Graphs

Essential Understanding: The Cartesian Coordinate System is a fundamental tool in mathematics that allows us to locate points on a plane, graph relationships between quantities, and solve problems visually. Named after the French mathematician René Descartes, this system forms the foundation for understanding linear functions, coordinate geometry, and much more.

X-Axis: Horizontal axis
Y-Axis: Vertical axis
Origin: (0, 0)

Understanding the Cartesian Plane

Coordinates

An ordered pair of numbers \((x, y)\) that specifies the location of a point on the plane.

x = horizontal position

y = vertical position

Think: "Crawl before you climb"

The Axes

X-axis: Horizontal number line

Y-axis: Vertical number line

Think: "X is a crossbar, Y goes up"

The Origin

The point where the axes intersect, denoted as (0, 0).

This is the starting point for all coordinate measurements.

Think: "All journeys begin at home"

Interactive Coordinate Visualizer

Click to Plot Points on the Coordinate Plane

Click anywhere on the grid to plot a point. The coordinates will snap to the nearest integer.
Click to plot
No point yet
Points plotted: 0

The Four Quadrants

The Cartesian plane is divided into four quadrants by the x-axis and y-axis. Understanding which quadrant a point lies in helps determine the signs of its coordinates.

Quadrant I

+ + P(x, y)

x > 0, y > 0

Both coordinates positive

Example: (3, 5)

Quadrant II

+ + P(x, y)

x < 0, y > 0

x negative, y positive

Example: (-3, 5)

Quadrant III

- + P(x, y)

x < 0, y < 0

Both coordinates negative

Example: (-3, -5)

Quadrant IV

+ - P(x, y)

x > 0, y < 0

x positive, y negative

Example: (3, -5)

Remembering Quadrants

  • "All Students Take Calculus" - Start from Quadrant I and go counterclockwise
  • Quadrant I: All positive
  • Quadrant II: Sine is positive (x negative, y positive)
  • Quadrant III: Tangent is positive (both negative)
  • Quadrant IV: Cosine is positive (x positive, y negative)

Plotting Points on the Cartesian Plane

To plot a point \((x, y)\) on the Cartesian plane, follow these simple steps:

1
Start at the Origin: Begin at (0, 0) where the x-axis and y-axis intersect.
2
Move along the x-axis: Move right if x is positive, or left if x is negative. This is called the "crawl" motion.
3
Move along the y-axis: Move up if y is positive, or down if y is negative. This is called the "climb" motion.
4
Mark the point: Place a dot at your final position and label it with its coordinates.
1

Worked Example: Plotting Points

Problem: Plot the following points on the Cartesian plane: A(2, 3), B(-4, 2), C(-3, -5), and D(5, -4)

X Y 2 4 -2 -4 -2 2 4 I II III IV O A(2, 3) B(-4, 2) C(-3, -5) D(5, -4)

Solution:

  • Point A(2, 3): From the origin, move 2 units right (positive x) and 3 units up (positive y) → Quadrant I
  • Point B(-4, 2): From the origin, move 4 units left (negative x) and 2 units up (positive y) → Quadrant II
  • Point C(-3, -5): From the origin, move 3 units left (negative x) and 5 units down (negative y) → Quadrant III
  • Point D(5, -4): From the origin, move 5 units right (positive x) and 4 units down (negative y) → Quadrant IV

Linear Functions and Their Graphs

A linear function is a function whose graph is a straight line. The general form is \(y = mx + c\), where:

  • \(m\) is the gradient (slope) of the line
  • \(c\) is the y-intercept (where the line crosses the y-axis)

Types of Linear Functions

Horizontal Line:
\[ y = c \]

Gradient = 0, parallel to x-axis

Vertical Line:
\[ x = k \]

Undefined gradient, parallel to y-axis

Linear Function:
\[ y = mx + c \]

General form with gradient m

Interactive Linear Graph Explorer

Explore Linear Functions

1
2

Equation: y = 1x + 2

Y-Intercept: (0, 2)

Finding Intercepts

Intercepts are points where a graph crosses the axes. They are important for graphing linear functions.

Intercept Definition How to Find
Y-Intercept Where the line crosses the y-axis Set \(x = 0\), solve for \(y\) → point \((0, c)\)
X-Intercept Where the line crosses the x-axis Set \(y = 0\), solve for \(x\) → point \((a, 0)\)
2

Worked Example: Finding Intercepts

Problem: Find the x-intercept and y-intercept of the line \(y = 2x - 4\).

1
Find the y-intercept:
Set \(x = 0\):
\[ y = 2(0) - 4 = -4 \]
The y-intercept is (0, -4)
2
Find the x-intercept:
Set \(y = 0\):
\[ 0 = 2x - 4 \]
\[ 2x = 4 \]
\[ x = 2 \]
The x-intercept is (2, 0)
3
Verify by graphing:
The line passes through (0, -4) on the y-axis and (2, 0) on the x-axis.

Gradient (Slope) of a Line

The gradient (or slope) measures how steep a line is. It is calculated as the ratio of the vertical change to the horizontal change between two points.

Gradient Formula

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

Where \(m\) is the gradient, and \((x_1, y_1)\) and \((x_2, y_2)\) are two points on the line.

A(1, 5) B(5, 2) Run = 4 Rise = -3 m = rise/run = (-3)/4 = -0.75

Types of Gradient

Positive Rising from left to right

Positive Gradient (m > 0)

Line rises from left to right

Example: y = 2x + 3

Negative Falling from left to right

Negative Gradient (m < 0)

Line falls from left to right

Example: y = -2x + 3

Zero Horizontal line

Zero Gradient (m = 0)

Horizontal line

Example: y = 5

Undefined Vertical line

Undefined Gradient

Vertical line

Example: x = 5

3

Worked Example: Calculating Gradient

Problem: Calculate the gradient of the line passing through points A(2, 3) and B(6, 11).

1
Identify the coordinates:
Point A: \((x_1, y_1) = (2, 3)\)
Point B: \((x_2, y_2) = (6, 11)\)
2
Apply the gradient formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2 \]
3
State the answer:
The gradient is 2. Since the gradient is positive, the line rises from left to right.

Past Paper Style Questions

CSEC-Style Question 1

(a) Write down the coordinates of the points shown on the grid below:

-4 -3 -2 -1 1 2 3 1 2 -1 -2 A B C D

(b) Which of these points lies in Quadrant III?

(c) Calculate the distance from A to B.

Solutions:

(a) A(-3, 2), B(2, -1), C(-2, -1), D(1, -2)

(b) Point C(-2, -1) lies in Quadrant III (both coordinates negative)

(c) Using the distance formula: \[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(2-(-3))^2 + (-1-2)^2} = \sqrt{5^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34} \]

CSEC-Style Question 2

(a) Find the equation of a line with gradient 3 passing through the point (0, 4).

(b) Find the x-intercept and y-intercept of the line \(y = 2x - 6\).

(c) Two points P(2, 5) and Q(6, 13) lie on a straight line. Find the gradient of PQ.

Solutions:

(a) Using y = mx + c: y = 3x + 4

(b) Y-intercept: x = 0, y = 2(0) - 6 = -6 → (0, -6)
X-intercept: y = 0, 0 = 2x - 6, 2x = 6, x = 3 → (3, 0)

(c) m = (13 - 5)/(6 - 2) = 8/4 = 2

CSEC-Style Question 3

The table below shows the relationship between x and y for a linear function.

x 1 2 3 4
y 3 5 7 9

(a) Write the equation relating x and y.

(b) What is the value of y when x = 0?

(c) What is the value of x when y = 15?

Solutions:

(a) The pattern shows y increases by 2 for each increase of 1 in x. The equation is y = 2x + 1

(b) When x = 0, y = 2(0) + 1 = 1

(c) When y = 15: 15 = 2x + 1, 2x = 14, x = 7

CSEC Practice Arena

Test Your Understanding

1
What are the coordinates of the point shown on the grid where the x-coordinate is -3 and the y-coordinate is 4?
(4, -3)
(-3, 4)
(3, 4)
(-3, -4)
Solution: The point has x = -3 and y = 4, so the coordinates are (-3, 4). This point is in Quadrant II since x is negative and y is positive.
2
Which quadrant contains the point (-5, -2)?
Quadrant I
Quadrant II
Quadrant III
Quadrant IV
Solution: Quadrant III contains points where both x and y are negative (-5, -2).
3
Find the y-intercept of the line y = -2x + 5.
x = -2
y = 2
(0, 5)
(5, 0)
Solution: The y-intercept is found by setting x = 0: y = -2(0) + 5 = 5. So the y-intercept is (0, 5).
4
Calculate the gradient of the line passing through (2, 3) and (6, 11).
1
2.5
2
4
Solution: m = (11 - 3)/(6 - 2) = 8/4 = 2
5
Which of the following equations represents a horizontal line?
y = 2x
x = 4
y = 4
y = x + 1
Solution: y = 4 is a horizontal line because the y-value is constant (gradient = 0). x = 4 is a vertical line.
Target

CSEC Examination Tips

  • Order matters: Coordinates are written as (x, y) - never swap them!
  • Quadrants: Remember "All Students Take Calculus" to recall the order of quadrants
  • Intercepts: Y-intercept when x = 0; X-intercept when y = 0
  • Gradient: Rise over run - always (change in y)/(change in x)
  • Positive gradient: Line goes up from left to right
  • Negative gradient: Line goes down from left to right
  • Graphing: Plot intercepts first, then use gradient to find more points
  • Check your answer: Verify that your plotted points satisfy the equation

Summary: Essential Concepts

Coordinate System

  • Origin: (0, 0)
  • X-axis: Horizontal
  • Y-axis: Vertical
  • Ordered pair: (x, y)

Quadrants

  • QI: (+, +)
  • QII: (-, +)
  • QIII: (-, -)
  • QIV: (+, -)

Linear Functions

  • Form: y = mx + c
  • Y-intercept: c
  • Gradient: m

Key Formulas

  • Gradient: m = (y₂-y₁)/(x₂-x₁)
  • Y-intercept: Set x = 0
  • X-intercept: Set y = 0
Scroll to Top