Introduction to Graphing
Learn how to read, interpret, and plot points on the coordinate plane — the foundation of all graphing in algebra.
🎯 By the end of this lesson you will be able to:
- Identify the parts of the coordinate plane (axes, origin, quadrants)
- Plot and read ordered pairs \((x, y)\)
- Determine which quadrant a point belongs to
- Calculate the distance between two points
- Find the midpoint between two points
📋 In This Lesson
1. The Coordinate Plane
💡 What is the Coordinate Plane?
The coordinate plane (also called the Cartesian plane) is a flat surface formed by two perpendicular number lines:
- The x-axis — runs horizontally (left ↔ right)
- The y-axis — runs vertically (up ↕ down)
- The origin — the point \((0, 0)\) where the axes cross
Every point on the plane is described by an ordered pair \((x, y)\), where \(x\) tells you how far to move left or right, and \(y\) tells you how far to move up or down.
🟢 Key Vocabulary
- Ordered pair: \((x, y)\) — always x first, then y
- x-coordinate: horizontal position (also called the abscissa)
- y-coordinate: vertical position (also called the ordinate)
- Origin: \((0, 0)\) — the center of the plane
🎬 Watch: The Coordinate Plane — plotting points and reading ordered pairs
2. Plotting Ordered Pairs
💡 How to Plot a Point
To plot the point \((3, -2)\):
- Start at the origin \((0,0)\)
- Move 3 units right along the x-axis (positive = right, negative = left)
- Move 2 units down along the y-axis (positive = up, negative = down)
- Place your point and label it
✏️ Worked Examples
Example 1. Plot and describe the point \(A = (4, 3)\)
👁️ Show Solution
Start at the origin. Move 4 units right (x = 4), then 3 units up (y = 3). Point A is in Quadrant I.
Example 2. Plot and describe the point \(B = (-2, 5)\)
👁️ Show Solution
Start at the origin. Move 2 units left (x = −2), then 5 units up (y = 5). Point B is in Quadrant II.
Example 3. Plot and describe the point \(C = (-3, -4)\)
👁️ Show Solution
Start at the origin. Move 3 units left (x = −3), then 4 units down (y = −4). Point C is in Quadrant III.
Example 4. What is special about the point \(D = (0, 5)\)?
👁️ Show Solution
Since \(x = 0\), the point lies directly on the y-axis. It is not in any quadrant.
3. The Four Quadrants
🟢 Quadrant Sign Rules
The axes divide the plane into four quadrants. The signs of \(x\) and \(y\) tell you exactly which quadrant a point is in:
- Quadrant I — \((+, +)\) — upper right
- Quadrant II — \((-, +)\) — upper left
- Quadrant III — \((-, -)\) — lower left
- Quadrant IV — \((+, -)\) — lower right
Points on an axis (where x = 0 or y = 0) are not in any quadrant.
✏️ Practice Problems
1. Which quadrant contains the point \((-5, 2)\)?
👁️ Show Solution
\(x\) is negative, \(y\) is positive → Quadrant II
2. Which quadrant contains the point \((7, -3)\)?
👁️ Show Solution
\(x\) is positive, \(y\) is negative → Quadrant IV
3. A point has a negative x-coordinate and a negative y-coordinate. Where is it?
👁️ Show Solution
Both negative → Quadrant III
4. Where is the point \((0, -4)\) located?
👁️ Show Solution
Since \(x = 0\), the point is on the negative y-axis — not in any quadrant.
4. Distance Between Two Points
💡 Where Does This Formula Come From?
The distance formula is just the Pythagorean theorem in disguise. If you draw a right triangle between two points, the distance between them is the hypotenuse:
$$a^2 + b^2 = c^2$$
Where \(a\) is the horizontal distance and \(b\) is the vertical distance.
🟢 The Distance Formula
For two points \((x_1, y_1)\) and \((x_2, y_2)\):
$$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
✏️ Worked Examples
Example 1. Find the distance between \((1, 2)\) and \((4, 6)\).
👁️ Show Solution
$$d = \sqrt{(4-1)^2 + (6-2)^2}$$
$$d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
The distance is 5 units.
Example 2. Find the distance between \((-3, 1)\) and \((2, -2)\).
👁️ Show Solution
$$d = \sqrt{(2-(-3))^2 + (-2-1)^2}$$
$$d = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$$
The distance is \(\sqrt{34} \approx 5.83\) units.
Practice 1. Find the distance between \((0, 0)\) and \((6, 8)\).
👁️ Show Solution
$$d = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
The distance is 10 units.
Practice 2. Find the distance between \((-1, -1)\) and \((3, 2)\).
👁️ Show Solution
$$d = \sqrt{(3-(-1))^2 + (2-(-1))^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$$
The distance is 5 units.
🎬 Watch: The Distance Formula — step by step with a full worked example
🎬 Watch: Distance Formula — a second worked example for extra practice
5. Midpoint Formula
💡 The Big Idea
The midpoint is the exact middle of a line segment. To find it, simply average the x-coordinates and average the y-coordinates of the two endpoints.
🟢 The Midpoint Formula
For two points \((x_1, y_1)\) and \((x_2, y_2)\):
$$M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$$
✏️ Worked Examples
Example 1. Find the midpoint of \((2, 4)\) and \((8, 10)\).
👁️ Show Solution
$$M = \left(\frac{2+8}{2},\ \frac{4+10}{2}\right) = \left(\frac{10}{2},\ \frac{14}{2}\right) = (5,\ 7)$$
The midpoint is (5, 7).
Example 2. Find the midpoint of \((-4, 1)\) and \((2, -5)\).
👁️ Show Solution
$$M = \left(\frac{-4+2}{2},\ \frac{1+(-5)}{2}\right) = \left(\frac{-2}{2},\ \frac{-4}{2}\right) = (-1,\ -2)$$
The midpoint is (−1, −2).
Practice 1. The midpoint of a segment is \((3, 5)\). One endpoint is \((1, 2)\). Find the other endpoint.
👁️ Show Solution
Use the midpoint formula in reverse. If \(M = (3,5)\) and one endpoint is \((1,2)\):
$$3 = \frac{1 + x_2}{2} \Rightarrow x_2 = 5$$
$$5 = \frac{2 + y_2}{2} \Rightarrow y_2 = 8$$
The other endpoint is (5, 8).
Practice 2. Find the midpoint of \((0, -6)\) and \((-4, 0)\).
👁️ Show Solution
$$M = \left(\frac{0+(-4)}{2},\ \frac{-6+0}{2}\right) = \left(-2,\ -3\right)$$
The midpoint is (−2, −3).