Functions: Coordinate System

Learn the two axes of a grid, the origin, and how to write and locate points.

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Learning Guide: The Coordinate System

The Axes & The Origin

The grid is built from two intersecting axes:

x-axis: The horizontal axis. Its values increase as you move right (positive direction) and decrease as you move left (negative direction).
y-axis: The vertical axis. Its values increase as you move upwards (positive direction) and decrease as you move downwards (negative direction).
Intersection: The two axes cross each other at a 90° angle (they are perpendicular). The point where they cross is called the origin, represented by coordinates $(0, 0)$ .

Figure 1: The coordinate plane showing the x-axis, y-axis, origin, the four quadrants, and a plotted point at (3, 2).

Writing points as Ordered Pairs

Any point on the coordinate plane is represented as an ordered pair written in the format:

(x, y)

The first number ( $x$ ) is the x-coordinate (abscissa). It tells you how many units to move horizontally from the origin (right if positive, left if negative).
The second number ( $y$ ) is the y-coordinate (ordinate). It tells you how many units to move vertically from the origin (up if positive, down if negative).

Coordinates of a Point

(x, y)

x: horizontal position
y: vertical position

The Four Quadrants (Regions)

The two crossing axes divide the coordinate system into four regions called quadrants, numbered counter-clockwise starting from the top-right:

First Quadrant (I): Top-right region. Both $x$ and $y$ are positive ( $x > 0, y > 0$ ). Example: $(2, 3)$ .
Second Quadrant (II): Top-left region. $x$ is negative and $y$ is positive ( $x < 0, y > 0$ ). Example: $(-4, 1)$ .
Third Quadrant (III): Bottom-left region. Both $x$ and $y$ are negative ( $x < 0, y < 0$ ). Example: $(-2, -3)$ .
Fourth Quadrant (IV): Bottom-right region. $x$ is positive and $y$ is negative ( $x > 0, y < 0$ ). Example: $(3, -5)$ .

Note: Points that lie directly on the x-axis or y-axis (like

(0, 4)

(-3, 0)

) or at the origin

(0, 0)

are not located in any of the four quadrants.

Midpoint of a Segment

The midpoint

M

of a segment connecting points

A(x_1, y_1)

and

B(x_2, y_2)

is the point that lies exactly halfway between them. The coordinates of the midpoint are given by the formula card below:

Midpoint coordinates of a segment

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

A: (x₁, y₁), B: (x₂, y₂)

Mathematical Derivation & Example:
To understand why this formula works, we can project the segment

AB

and its midpoint

M

onto the coordinate axes. This splits both the horizontal and vertical intervals into two equal parts:

1. Horizontal Projection ( $x$ -axis):
The horizontal interval spans from

x_1

x_2

. On our graph, this goes from

x_1 = -3

x_2 = 5

(a total distance of

8

units). The midpoint

M

projects to

x_M = 1

. Since the distance from

x_1

x_M

must equal the distance from

x_M

x_2

x_M - x_1 = x_2 - x_M

Using our graph coordinates:

1 - (-3) = 5 - 1 = 4\text{ units}

Solving algebraically for

x_M

2x_M = x_1 + x_2 \implies x_M = \frac{x_1 + x_2}{2}

Substituting the values yields:

x_M = \frac{-3 + 5}{2} = \frac{2}{2} = 1

2. Vertical Projection ( $y$ -axis):
The vertical interval spans from

y_1

y_2

. On our graph, this goes from

y_1 = -2

y_2 = 4

(a total distance of

6

units). The midpoint

M

projects to

y_M = 1

. Equalizing the distances:

y_M - y_1 = y_2 - y_M

Using our graph coordinates:

1 - (-2) = 4 - 1 = 3\text{ units}

Solving algebraically for

y_M

2y_M = y_1 + y_2 \implies y_M = \frac{y_1 + y_2}{2}

Substituting the values yields:

y_M = \frac{-2 + 4}{2} = \frac{2}{2} = 1

By combining these components, we find the exact coordinates of the midpoint:

M(1, 1)

The graph shows that projecting the segment onto the

x

and

y

axes divides the interval between coordinates into two equal parts:

x_M - x_1 = x_2 - x_M

and

y_M - y_1 = y_2 - y_M

Distance Between Two Points

The distance

d

between two points

A(x_1, y_1)

and

B(x_2, y_2)

is the length of the straight line segment connecting them. The distance formula is given by the formula card below:

Distance between two points

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

d: Distance, A(x₁, y₁), B(x₂, y₂)

Relation to the Pythagorean Theorem:
By projecting the segment

AB

onto the axes, we construct a right triangle

ACB

with a horizontal leg

AC

(length

|x_2 - x_1|

) and a vertical leg

CB

(length

|y_2 - y_1|

).
Applying the Pythagorean theorem (

a^2 + b^2 = c^2

), the hypotenuse

d

satisfies:

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

For example, using the points

A(-3, -2)

and

B(5, 4)

on the graph:
- The horizontal distance is

x_2 - x_1 = 5 - (-3) = 8

.
- The vertical distance is

y_2 - y_1 = 4 - (-2) = 6

.
- Applying the theorem:

d = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10

The graph shows the right triangle

ACB

formed by projecting the segment

AB

onto the axes, demonstrating how the distance formula is derived directly from the Pythagorean theorem.

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