Lesson 2.1: The Rectangular Coordinate Systems and Graphs

New Concept

The rectangular coordinate system is a powerful tool that connects algebra and geometry.
You'll learn to plot points, graph equations, find key features like intercepts, and calculate distances and midpoints in a two-dimensional plane.

What’s next

You're just getting started! Next, you'll tackle interactive examples on plotting points, finding intercepts, and applying the distance and midpoint formulas through practice cards.

Property

A two-dimensional plane where the x-axis is the horizontal axis and the y-axis is the vertical axis.
A point in the plane is defined as an ordered pair, $(x, y)$, such that $x$ is determined by its horizontal distance from the origin and $y$ is determined by its vertical distance from the origin.

Examples

• To plot the point $(-3, 5)$, you start at the origin $(0,0)$, move 3 units to the left along the x-axis, and then 5 units up, parallel to the y-axis.
• The point $(4, 0)$ is located by moving 4 units to the right from the origin. Since its y-coordinate is 0, it lies directly on the x-axis.
• The point $(2, -2)$ is found by starting at the origin, moving 2 units right, and then 2 units down. This point is in the fourth quadrant.

Explanation

Think of the Cartesian system as a map for numbers. The two axes, x (horizontal) and y (vertical), act like crossing roads. Any point on this map can be found using a unique address called an ordered pair, $(x, y)$.

Property

To graph an equation in two variables, such as $y = 2x - 1$, we can plot a set of points. First, make a table with columns for $x$, the equation, and the resulting $(x, y)$ ordered pair. Choose several x-values, calculate the corresponding y-values, plot these ordered pairs, and connect them if they form a line.

Examples

• To graph $y = x + 3$, pick x-values like $-1, 0, 1$. The points are $(-1, 2)$, $(0, 3)$, and $(1, 4)$. Plotting these points reveals a straight line.
• For the equation $y = -2x + 1$, if we choose $x=0$, we get $y=1$ for the point $(0, 1)$. If we choose $x=2$, we get $y=-3$ for the point $(2, -3)$. Connecting these gives the graph.
• To graph $y = \frac{1}{2}x - 1$, choose even numbers for $x$ to avoid fractions. For $x=-2, 0, 2$, we get the points $(-2, -2)$, $(0, -1)$, and $(2, 0)$.

Explanation

Graphing by plotting points is like connecting the dots to reveal a picture of an equation. By choosing a few x-values and finding their y-partners, you create guideposts that show the shape and location of the equation's graph.

Property

The intercepts of a graph are points at which the graph crosses the axes. The x-intercept is the point where the graph crosses the x-axis, and its y-coordinate is always zero. The y-intercept is where the graph crosses the y-axis, and its x-coordinate is always zero.
To find the intercepts from an equation:

• Find the x-intercept by setting $y = 0$ and solving for $x$.
• Find the y-intercept by setting $x = 0$ and solving for $y$.

Examples

• For the equation $y = 2x - 6$: Set $y=0$ to get the x-intercept at $(3, 0)$. Set $x=0$ to get the y-intercept at $(0, -6)$.
• In $2x + 3y = 12$: Set $y=0$, then $2x=12$, so the x-intercept is $(6, 0)$. Set $x=0$, then $3y=12$, so the y-intercept is $(0, 4)$.
• To find the intercepts of $y = -x + 5$: Set $y=0$ to find the x-intercept $(5, 0)$. Set $x=0$ to find the y-intercept $(0, 5)$.

Explanation

Intercepts are special points where a graph hits the main axes. The x-intercept is where the line crosses the horizontal x-axis ($y=0$), and the y-intercept is where it crosses the vertical y-axis ($x=0$). They are useful landmarks for graphing.

Property

Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. Given endpoints $(x_1, y_1)$ and $(x_2, y_2)$, the distance $d$ between them is given by:

Examples

• To find the distance between $(1, 2)$ and $(4, 6)$, use the formula: $d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$.
• The distance between $(-2, 5)$ and $(3, -1)$ is $d = \sqrt{(3 - (-2))^2 + (-1 - 5)^2} = \sqrt{5^2 + (-6)^2} = \sqrt{25+36} = \sqrt{61}$.
• The distance between $(-4, 3)$ and $(5, 3)$ is $d = \sqrt{(5 - (-4))^2 + (3 - 3)^2} = \sqrt{9^2 + 0^2} = \sqrt{81} = 9$.

Explanation

The distance formula calculates the straight-line distance between any two points on a graph. It works by creating a right triangle with the line segment as the hypotenuse and using the Pythagorean theorem to find its length.

Property

When the endpoints of a line segment are known, we can find the point midway between them using the midpoint formula. Given the endpoints $(x_1, y_1)$ and $(x_2, y_2)$, the coordinates of the midpoint $M$ are found with the formula:

Examples

• The midpoint of the segment with endpoints $(2, 4)$ and $(6, 8)$ is found by $(\frac{2+6}{2}, \frac{4+8}{2}) = (4, 6)$.
• To find the midpoint between $(-3, 5)$ and $(1, -1)$, calculate $(\frac{-3+1}{2}, \frac{5+(-1)}{2}) = (-1, 2)$.
• The center of a circle with a diameter having endpoints $(-4, -2)$ and $(6, 0)$ is the midpoint: $(\frac{-4+6}{2}, \frac{-2+0}{2}) = (1, -1)$.

Explanation

The midpoint formula gives you the exact center of a line segment. It works by finding the average of the x-coordinates and the average of the y-coordinates of the two endpoints, giving you the coordinates of the middle point.