Distance and Midpoint Formulas in the Cartesian Plane

Chapter 2: Graphs

Section 2.1: The Distance and Midpoint Formulas

This section introduces the fundamental concepts of the rectangular (Cartesian) coordinate system and explains how to calculate the distance and midpoint between two points in the plane. These tools are essential for analyzing geometric relationships and graphing in precalculus.

Rectangular (Cartesian) Coordinates

• Definition: The rectangular coordinate system consists of two real number lines intersecting at right angles: the x-axis (horizontal) and the y-axis (vertical). Their intersection is called the origin (O).

• The plane formed by these axes is called the xy-plane, and the axes themselves are the coordinate axes.

• Any point P in the xy-plane is represented by an ordered pair (x, y), where x is the x-coordinate (abscissa) and y is the y-coordinate (ordinate).

• The coordinate axes divide the plane into four quadrants:

  • Quadrant I: x > 0, y > 0

  • Quadrant II: x < 0, y > 0

  • Quadrant III: x < 0, y < 0

  • Quadrant IV: x > 0, y < 0

Distance Formula

The distance formula allows you to find the length between two points in the plane using their coordinates.

• Theorem 1 (Distance Formula): The distance d between points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ is: $d(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Application: This formula is derived from the Pythagorean Theorem, treating the difference in x and y as the legs of a right triangle.

Example: Find the distance between (0, 2) and (4, 5): $\begin{align*} d &= \sqrt{(4 - 0)^2 + (5 - 2)^2} \\ &= \sqrt{16 + 9} \\ &= \sqrt{25} = 5 \end{align*}$

Example: Find the distance between (-4, 5) and (1, 3): $\begin{align*} d &= \sqrt{(1 - (-4))^2 + (3 - 5)^2} \\ &= \sqrt{5^2 + (-2)^2} \\ &= \sqrt{25 + 4} \\ &= \sqrt{29} \approx 5.39 \end{align*}$

Using Algebra to Solve Geometry Problems

Algebraic methods can be used to analyze geometric figures in the coordinate plane, such as triangles.

• Finding Side Lengths: Use the distance formula for each pair of vertices.

• Verifying Right Triangles: If the sum of the squares of two sides equals the square of the third side, the triangle is a right triangle (by the converse of the Pythagorean Theorem).

• Finding Area: For a right triangle, area is half the product of the lengths of the legs forming the right angle.

Example: Given points $A = (-4, 1)$, $B = (3, 4)$, $C = (4, 1)$:

• Find side lengths:

  • $d(A, B) = \sqrt{(3 - (-4))^2 + (4 - 1)^2} = \sqrt{49 + 9} = \sqrt{58}$

  • $d(B, C) = \sqrt{(4 - 3)^2 + (1 - 4)^2} = \sqrt{1 + 9} = \sqrt{10}$

  • $d(A, C) = \sqrt{(4 - (-4))^2 + (1 - 1)^2} = \sqrt{64 + 0} = 8$

• Verify right triangle:

  • If $d(A, B)^2 + d(A, C)^2 = d(B, C)^2$, then triangle is right.

• Find area:

  • $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$

Midpoint Formula

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

• Theorem 2 (Midpoint Formula): The midpoint $M = (x, y)$ of the segment from $P_1 = (x_1, y_1)$ to $P_2 = (x_2, y_2)$ is: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Example: Find the midpoint between $(-4, 4)$ and $(4, 2)$: $\begin{align*} x &= \frac{-4 + 4}{2} = 0 \\ y &= \frac{4 + 2}{2} = 3 \\ M &= (0, 3) \end{align*}$

Summary Table: Key Formulas

Concept	Formula	Description
Distance between $(x_1, y_1)$ and $(x_2, y_2)$	$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$	Length of segment connecting two points
Midpoint between $(x_1, y_1)$ and $(x_2, y_2)$	$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$	Coordinates of the point halfway between two points
Area of a right triangle	$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$	Area when base and height are known

Additional info: These formulas are foundational for analytic geometry and are frequently used in graphing, geometry, and further studies in calculus and physics.