MATH 140 Section 1.1: Rectangular Coordinates

Course: MATH 140 - PRECALCULUS
Unit: 1. Functions and Their Graphs
Section: 1.1 Rectangular Coordinates
Institution: Harrisburg University of Science and Technology
Instructor: Dr. Devi Bhakta Suberi
Core Learning Objectives:
- Plot points in the Cartesian plane.
- Use the Distance Formula to find the distance between two points.
- Use the Midpoint Formula to find the midpoint of a line segment.
- Use a coordinate plane to model and solve real-life problems.

Definition and Structure:
- Real numbers can be represented by points on a real number line, while ordered pairs of real numbers are represented by points in a two-dimensional plane called the rectangular coordinate system, or the Cartesian plane.
- The system is named after the French mathematician René Descartes ( $1596\text{--}1650$ ).
- The Cartesian plane is constructed by two real number lines intersecting at right angles.
- The horizontal real number line is called the $x\text{-axis}$ .
- The vertical real number line is called the $y\text{-axis}$ .
- The point where the two axes intersect is called the origin, represented as $(0, 0)$ .
- The intersecting axes divide the plane into four regions called quadrants: Quadrant I, Quadrant II, Quadrant III, and Quadrant IV.

Coordinates of a Point:
- Every point in the plane corresponds to a unique ordered pair $(x, y)$ of real numbers $x$ and $y$ , known as the coordinates of the point.
- The $x\text{-coordinate}$ represents the directed distance from the $y\text{-axis}$ to the point.
- The $y\text{-coordinate}$ represents the directed distance from the $x\text{-axis}$ to the point.
- Notation Note: The notation $(x, y)$ is used to denote both a point in the coordinate plane and an open interval on the real number line. Context determines which interpretation is intended.

Significance:
- The rectangular coordinate system provides a method to visualize relationships between two variables.
- Descartes's introduction of coordinates in the plane is foundational and widely applied across scientific and business-related disciplines.
Example 1: Plotting Points in the Cartesian Plane
- Problem: Plot the points $(-1, 2)$ , $(3, 4)$ , $(0, 0)$ , $(3, 0)$ , and $(-2, -3)$ .
- Procedure:
- To plot the point $(-1, 2)$ , construct a vertical line through $-1$ on the $x\text{-axis}$ and a horizontal line through $2$ on the $y\text{-axis}$ .
- The point of intersection of these two lines is $(-1, 2)$ .
- Apply equivalent steps for the remaining ordered pairs.

The Pythagorean Theorem:
- For a right triangle with legs of lengths $a$ and $b$ and a hypotenuse of length $c$ : $a^2 + b^2 = c^2$
- Converse: If a triangle has side lengths $a$ , $b$ , and $c$ such that $a^2 + b^2 = c^2$ , then the triangle is a right triangle.

Derivation of the Distance Formula:
- Consider two points $(x_1, y_1)$ and $(x_2, y_2)$ in the Cartesian plane.
- A right triangle is formed using $(x_1, y_1)$ and $(x_2, y_2)$ as the acute vertices and $(x_1, y_2)$ as the right-angled vertex.
- The length of the horizontal side of the triangle is given by the absolute difference $|x_2 - x_1|$ .
- The length of the vertical side of the triangle is given by the absolute difference $|y_2 - y_1|$ .
- Let $d$ be the length of the hypotenuse, which represents the distance between $(x_1, y_1)$ and $(x_2, y_2)$ .
- Applying the Pythagorean Theorem: $d^2 = |x_2 - x_1|^2 + |y_2 - y_1|^2$ $d = \sqrt{|x_2 - x_1|^2 + |y_2 - y_1|^2}$ $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The Distance Formula:
- The distance $d$ between any two points $(x_1, y_1)$ and $(x_2, y_2)$ in the plane is: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Example: Finding the Distance Between Two Points
- Problem: Find the distance between the points $(-2, 1)$ and $(3, 4)$ .
- Solution:
- Assign $(x_1, y_1) = (-2, 1)$ and $(x_2, y_2) = (3, 4)$ .
- Substitute the values into the Distance Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $d = \sqrt{[3 - (-2)]^2 + (4 - 1)^2}$ $d = \sqrt{(5)^2 + (3)^2}$ $d = \sqrt{25 + 9}$ $d = \sqrt{34}$ $d \approx 5.83$
- The distance between the points is approximately $5.83\,\text{units}$ .

Definition and Concept:
- The midpoint of a line segment is the point that divides the segment into two equal parts.
- To find the midpoint of a line segment connecting two points in a coordinate plane, calculate the average values of the $x\text{-coordinates}$ and the average values of the $y\text{-coordinates}$ of the endpoints.
The Midpoint Formula:
- The midpoint of the line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is: $\text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
Example: Finding a Line Segment's Midpoint
- Problem: Find the midpoint of the line segment joining the points $(-5, -3)$ and $(9, 3)$ .
- Solution:
- Let $(x_1, y_1) = (-5, -3)$ and $(x_2, y_2) = (9, 3)$ .
- Substitute the coordinates into the Midpoint Formula: $\text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ $\text{Midpoint} = \left(\frac{-5 + 9}{2}, \frac{-3 + 3}{2}\right)$ $\text{Midpoint} = \left(\frac{4}{2}, \frac{0}{2}\right)$ $\text{Midpoint} = (2, 0)$
- The midpoint of the line segment is $(2, 0)$ .

Coordinate Modeling:
- Real-life spatial problems can be modeled on a coordinate plane by assigning coordinates to positions, allowing geometric formulas like distance and midpoint to solve practical problems.
Example: Finding the Length of a Football Pass
- Scenario: A quarterback throws a football pass from the $28\text{-yard}$ line, $40\,\text{yards}$ from the sideline. A wide receiver catches the pass on the $5\text{-yard}$ line, $20\,\text{yards}$ from the same sideline.
- Coordinate Assignment:
- Quarterback position: $(40, 28)$
- Receiver position: $(20, 5)$

Calculation:
- Find the length of the pass by computing the distance between $(40, 28)$ and $(20, 5)$ : $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $d = \sqrt{(40 - 20)^2 + (28 - 5)^2}$ $d = \sqrt{(20)^2 + (23)^2}$ $d = \sqrt{400 + 529}$ $d = \sqrt{929}$ $d \approx 30$
Conclusion: The pass is approximately $30\,\text{yards}$ long.