Grade 10

Grade 10Coordinate Geometry


Area of a Triangle


In coordinate geometry, we often deal with shapes on the plane, which means they will have points located in their coordinate system. One of the fundamental shapes in geometry is the triangle. Learning how to find the area of a triangle, especially when it is placed in the coordinate plane, is an important step in understanding spatial relationships and calculations in mathematics.

Let us delve deeper into this topic and understand how we can use the coordinates of the vertices of a triangle to determine its area.

Coordinate geometry basics

Before we learn the area of a triangle, it is necessary to recall some basic concepts of coordinate geometry:

  • The coordinate plane is made up of two perpendicular lines - the x-axis (horizontal) and the y-axis (vertical).
  • Any point in this plane can be represented as a pair of coordinates (x, y), where 'x' is the horizontal distance from the origin, and 'y' is the vertical distance from the origin.

The formula for the area of a triangle in the coordinate plane

Suppose you have a triangle whose vertices are located at ( A(x_1, y_1) ), ( B(x_2, y_2) ), and ( C(x_3, y_3) ). The formula to calculate the area of the triangle formed by these points is:

        text{Area} = frac{1}{2} left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) right|

Let us understand this formula:

  • The vertical bars (| cdot |) represent the absolute value, which ensures that the area is always a positive number, since the physical space occupied cannot be negative.
  • The expression inside the absolute value reduces when specific axes and points are considered.
  • The formula itself comes from the determinant of the matrix. It is a method used with vectors and determinants that simplifies these calculations.

Visualizing with examples

Looking at a triangle on the coordinate plane helps to understand how this formula works. Consider the following example:

Example 1: Simple triangular area calculation

Given a triangle whose vertices are ( A(2, 3) ), ( B(5, 11) ), and ( C(9, 7) ), find the area of it.

A(2,3) b(5,11) C(9,7)

Using the area formula:

        text{Area} = frac{1}{2} left| 2(11-7) + 5(7-3) + 9(3-11) right|
        text{Area} = frac{1}{2} left| 2 times 4 + 5 times 4 + 9 times (-8) right|
        text{Area} = frac{1}{2} left| 8 + 20 - 72 right|
        text{Area} = frac{1}{2} left| -44 right|
        text{Area} = frac{1}{2} times 44 
        text{Area} = 22 text{ square units}

Example 2: Collinear points

What if the points of the triangle are collinear (i.e. they all lie on a straight line)? Consider the points ( A(1, 2) ), ( B(3, 4) ), and ( C(5, 6) ).

A(1,2) b(3,4) C(5,6)

Using the area formula:

        text{Area} = frac{1}{2} left| 1(4-6) + 3(6-2) + 5(2-4) right|
        text{Area} = frac{1}{2} left| 1 times (-2) + 3 times 4 + 5 times (-2) right|
        text{Area} = frac{1}{2} left| -2 + 12 - 10 right|
        text{Area} = frac{1}{2} left| 0 right|
        text{Area} = 0 text{ square units}

The area is zero because the points are collinear, thus not forming a triangle in the traditional sense with actual surface area.

Applications and significance

Calculating the area of a triangle in coordinate geometry is useful in a variety of fields, such as computer graphics, geolocation computations, and various scientific disciplines where location and structure need to be quantified.

This concept also strengthens the connection between algebra and geometry, and shows how equations can represent spatial relationships.

Practice problems

Solve the following practice problems to strengthen your understanding:

  1. Find the area of a triangle whose vertices are ( A(0,0) ), ( B(6,0) ), and ( C(6,8) ).
  2. Find the area of a triangle with vertices ( A(-3,7) ), ( B(3,7) ), and ( C(0,-2) ).
  3. Determine the area of the triangle at the points ( A(-1,-1) ), ( B(2,3) ), and ( C(4,0) ).

Solutions to practice problems

After you've solved the problems, compare your solution to the steps below:

  1. Vertices ( A(0,0) ), ( B(6,0) ), and ( C(6,8) ):
                    text{Area} = frac{1}{2} left| 0(0-8) + 6(8-0) + 6(0-0) right|
                    text{Area} = frac{1}{2} left| 0 + 48 + 0 right|
                    text{Area} = frac{1}{2} times 48 
                    text{Area} = 24 text{ square units}
            
  2. Vertex ( A(-3,7) ), ( B(3,7) ), ( C(0,-2) ):
                    text{Area} = frac{1}{2} left| -3(7+2) + 3(-2-7) + 0(7-7) right|
                    text{Area} = frac{1}{2} left| -27 - 27 + 0 right|
                    text{Area} = frac{1}{2} times 54
                    text{Area} = 27 text{ square units}
            
  3. Vertex ( A(-1,-1) ), ( B(2,3) ), ( C(4,0) ):
                    text{Area} = frac{1}{2} left| -1(3-0) + 2(0+1) + 4(-1-3) right|
                    text{Area} = frac{1}{2} left| -3 + 2 - 16 right|
                    text{Area} = frac{1}{2} times 17
                    text{Area} = 8.5 text{ square units}
            

As you practice further, you will become more familiar with manipulating coordinates and using determinant-based formulas intuitively. With consistent practice and understanding, these calculations will become second nature.


Grade 10 → 3.4


U
username
0%
completed in Grade 10


Comments