Examples of Using Coordinate Geometry

Explore practical examples of using coordinate geometry to solve geometric problems effectively.
By Jamie

Using Coordinate Geometry to Solve Geometric Problems

Coordinate geometry, or analytic geometry, is a powerful tool for solving geometric problems by utilizing algebra and a coordinate system. This method allows us to translate geometric shapes into mathematical equations, making it easier to analyze and solve problems. Here, we present three practical examples that illustrate how coordinate geometry can be applied in various contexts.

Example 1: Finding the Distance Between Two Points

Context

In real-world applications, you might need to calculate the distance between two locations on a map. This example shows how to use the distance formula derived from coordinate geometry.

To find the distance between two points

  • Point A (2, 3)
  • Point B (7, 1)

The distance formula is given by:

D = √[(x₂ - x₁)² + (y₂ - y₁)²]

Substituting the coordinates:

  • x₁ = 2, y₁ = 3
  • x₂ = 7, y₂ = 1

D = √[(7 - 2)² + (1 - 3)²]
D = √[(5)² + (-2)²]
D = √[25 + 4]
D = √29 ≈ 5.39

Notes

This method can be extended to calculate the distance between multiple points or to find the midpoint of a line segment by using the midpoint formula, which averages the x and y coordinates of the endpoints.

Example 2: Determining the Equation of a Line

Context

Suppose you need to find the equation of a line that passes through two given points, which can help in various fields like engineering and economics. Here, we will find the equation of a line that goes through Points A (1, 2) and B (4, 6).

First, calculate the slope (m) of the line using the formula:

m = (y₂ - y₁) / (x₂ - x₁)

Substituting the coordinates:

  • y₁ = 2, y₂ = 6
  • x₁ = 1, x₂ = 4

m = (6 - 2) / (4 - 1) = 4 / 3

Next, use the point-slope form of the line equation:

y - y₁ = m(x - x₁)

Substituting Point A (1, 2):

y - 2 = (4/3)(x - 1)

Distributing the slope:

y - 2 = (4/3)x - (4/3)

Adding 2 to both sides gives the slope-intercept form:

y = (4/3)x + (2/3)

Notes

This example can be varied by changing the coordinates of the points or by finding the line perpendicular to this line, which can be useful in design and architecture.

Example 3: Finding the Area of a Triangle

Context

In many practical situations, such as land surveying or architecture, determining the area of a triangle formed by three vertices in a coordinate plane can be essential. Let’s find the area of a triangle formed by the points A (1, 2), B (4, 6), and C (5, 1).

To calculate the area (A) of a triangle given vertices A, B, and C, use the formula:

A = (1/2) | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |

Substituting the coordinates:

  • A (1, 2), B (4, 6), C (5, 1)

A = (1/2) | 1(6 - 1) + 4(1 - 2) + 5(2 - 6) |
A = (1/2) | 1(5) + 4(-1) + 5(-4) |
A = (1/2) | 5 - 4 - 20 |
A = (1/2) | -19 |
A = 9.5

Notes

This technique can be adapted for finding the area of more complex shapes by subdividing them into triangles, and it can also be used to determine whether three points are collinear by checking if the area equals zero.