Equations, Inequalities, and Modeling: The Cartesian Coordinate System, Pythagorean Theorem, Distance and Midpoint Formulas

Equations, Inequalities, and Modeling

The Cartesian Coordinate System

The Cartesian coordinate system, also known as the rectangular coordinate system, is a two-dimensional plane used to study the relationship between pairs of variables. It consists of a horizontal axis (x-axis) and a vertical axis (y-axis) that intersect at the origin (0,0). Each point in the plane is represented by an ordered pair (x, y).

• x-axis: The horizontal axis.

• y-axis: The vertical axis.

• Ordered pair (x, y): Represents a point where x is the horizontal position and y is the vertical position.

• Example: The point (3, 11) means x = 3 and y = 11. This could represent, for example, a three-topping pizza costing $11.

• To plot a point, locate the x-value on the x-axis and the y-value on the y-axis, then find where they intersect.

Note: Improper fractions should be converted to mixed numbers or decimals before plotting.

The Pythagorean Theorem

The Pythagorean Theorem is a fundamental relation in geometry among the three sides of a right triangle. It states that the sum of the squares of the two legs is equal to the square of the hypotenuse (the side opposite the right angle).

• Formula:

$a^2 + b^2 = c^2$

• a, b: The lengths of the legs of the right triangle.

• c: The length of the hypotenuse.

• Example: If a right triangle has legs of lengths 3 and 4, the hypotenuse is $c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

Application Example: If a 50-inch TV has a height of 24.5 inches, you can use the Pythagorean theorem to find the width if the diagonal (hypotenuse) is 50 inches:

$24.5^2 + w^2 = 50^2$

Solve for w:

$w^2 = 50^2 - 24.5^2$

$w = \sqrt{2500 - 600.25} = \sqrt{1899.75} \approx 43.6$ inches

The Distance Formula

The distance formula is used to calculate the distance between two points in the Cartesian plane. It is derived from the Pythagorean theorem.

• Formula:

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

• (x_1, y_1) and (x_2, y_2): The coordinates of the two points.

• Example: To find the distance between (-1, 1) and (-6, 5):

$d = \sqrt{(-6 - (-1))^2 + (5 - 1)^2} = \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \approx 6.4$

The Midpoint Formula

The midpoint formula finds the point exactly halfway between two given points in the coordinate plane.

• Formula:

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

• (x_1, y_1) and (x_2, y_2): The coordinates of the two endpoints.

• Example: The midpoint between (20.8, 1970) and (26.1, 2012) is:

$M = \left( \frac{20.8 + 26.1}{2}, \frac{1970 + 2012}{2} \right) = (23.45, 1991)$

Application: This can be used to find the average value or central point between two data points, such as the median age at first marriage for women in two different years.