Distance Formula

7.2 Distance Formula

Description

Quick Overview

The Distance Formula helps to calculate the distance between two points in a coordinate plane using their coordinates.

Standard

This section introduces the Distance Formula, derived from the Pythagorean theorem. It explains how to find the distance between two points on the Cartesian plane, regardless of their location in relation to the axes. The formula is essential for various applications in geometry and real life.

Detailed

Distance Formula

In this section, we explore the Distance Formula, an essential tool in coordinate geometry used to calculate the distance between two points in a Cartesian coordinate system. The process involves applying the Pythagorean Theorem to the coordinates of the points in question.

Key Concepts Covered:

  1. Understanding Points on Axes: The section begins with visualizing points on the x-axis and y-axis, explaining how distances can simply be calculated by subtraction when points are aligned along the same axis.
  2. Using Pythagorean Theorem: When points are not aligned, the Distance Formula is derived by forming a right triangle from the points and applying the Pythagorean theorem. The formula can be expressed as:

\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

This formula allows us to find the distance between any two points, (x1, y1) and (x2, y2), in a two-dimensional space.

  1. Examples and Applications: The chapter demonstrates several examples, including scenarios like determining whether points form a triangle or calculating distances between specific points on a grid.
  2. Special Cases: Special cases are discussed, such as finding the distance from the origin and the unconditional distances between various geometric figures.

By the end of the section, students should be able to apply the Distance Formula effectively, whether for academic exercises or practical applications within fields such as engineering and physics.

Key Concepts

  • Understanding Points on Axes: The section begins with visualizing points on the x-axis and y-axis, explaining how distances can simply be calculated by subtraction when points are aligned along the same axis.

  • Using Pythagorean Theorem: When points are not aligned, the Distance Formula is derived by forming a right triangle from the points and applying the Pythagorean theorem. The formula can be expressed as:

  • \[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

  • This formula allows us to find the distance between any two points, (x1, y1) and (x2, y2), in a two-dimensional space.

  • Examples and Applications: The chapter demonstrates several examples, including scenarios like determining whether points form a triangle or calculating distances between specific points on a grid.

  • Special Cases: Special cases are discussed, such as finding the distance from the origin and the unconditional distances between various geometric figures.

  • By the end of the section, students should be able to apply the Distance Formula effectively, whether for academic exercises or practical applications within fields such as engineering and physics.

Memory Aids

🎵 Rhymes Time

  • In the plane, find the spaces, between the points, use their places.

📖 Fascinating Stories

  • Once there were two friends named Point A and Point B, who wanted to measure the distance between them without a ruler. They laid a right triangle down, showing how far apart they were, thanks to the Pythagorean theorem, they calculated their distance and became even closer friends.

🧠 Other Memory Gems

  • Base and Height for PQ; Distance is always the positive view.

🎯 Super Acronyms

D for Distance, P for Points, C for Calculate

  • DPC.

Examples

  • Example 1: Finding the distance between points A(4, 0) and B(6, 0) results in a distance of 2 units.

  • Example 2: Verifying if points P(3, 2), Q(-2, -3), and R(2, 3) form a triangle using the Distance Formula.

Glossary of Terms

  • Term: Distance Formula

    Definition:

    A formula used to determine the distance between two points (x1, y1) and (x2, y2) in the Cartesian plane: \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

  • Term: Pythagorean Theorem

    Definition:

    A mathematical principle that relates the lengths of the sides of a right triangle, defined as a² + b² = c².

  • Term: Collinear

    Definition:

    Points that lie on the same straight line.

  • Term: Quadrant

    Definition:

    One of the four regions created by the intersection of the x-axis and y-axis in a coordinate plane.

Example 4 - Variation

Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(2, 4)$ and $(6, 8)$.

Solution: Let $(x, y)$ be equidistant from the points $A(2, 4)$ and $B(6, 8)$.
We can set up the equation based on the distance formula:
\[ AP = BP \]
This gives us:
\[ \sqrt{(x - 2)^2 + (y - 4)^2} = \sqrt{(x - 6)^2 + (y - 8)^2} \]
Squaring both sides leads to:
\[ (x - 2)^2 + (y - 4)^2 = (x - 6)^2 + (y - 8)^2 \]
Expanding both sides results in:
\[ (x^2 - 4x + 4 + y^2 - 8y + 16) = (x^2 - 12x + 36 + y^2 - 16y + 64) \]
Thus, simplifying gives:
\[ -4x + 20 - 8y = -12x + 100 - 16y \]
This leads to the simplified equation:
\[ 8x - 8y = 80 \]
Hence, we find:
\[ x - y = 10 \]

Thus, the relation between $x$ and $y$ is $x - y = 10$.

Similar Question

Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(3, 1)$ and $(7, 5)$.

Solution: Let $(x, y)$ be equidistant from the points $A(3, 1)$ and $B(7, 5)$.
We have:
\[ AP = BP \]
Using the distance formula:
\[ \sqrt{(x - 3)^2 + (y - 1)^2} = \sqrt{(x - 7)^2 + (y - 5)^2} \]
Squaring both sides,
\[ (x - 3)^2 + (y - 1)^2 = (x - 7)^2 + (y - 5)^2 \]
Expanding and simplifying leads us to the relation.