Area of a Triangle in Coordinate Geometry

Coordinate geometry is defined as the study of geometry using the coordinate points on the plane with any dimension. Using coordinate geometry, it is possible to find the distance between two points, divide lines in a ratio, find the mid-point of a line, calculate the area of a triangle in the Cartesian plane etc.

There are various methods to find the area of the triangle according to the parameters given, like the base and height of the triangle, coordinates of vertices, length of sides, etc. In this article, we will discuss the method of finding area of any triangle when its coordinates are given.

In this method, if coordinates of the vertices of a triangle are given then we will see how to found the area of the triangle.

If coordinates of the triangle are (x₁, y₁), (x₂, y₂), and (x₃, y₃) then the area of the triangle is given by:

Derivation of the Formula

Given triangle PQR with coordinates P(x₁, y₁), Q(x₂, y₂), and R(x₃, y₃), lets find the formula for its area:

Step 1: Draw the perpendiculars from coordinates P, Q, and R to X-axis at A, B, and C respectively.

Step 2: Now if we look at the figure carefully, three different trapeziums are formed such as PQAB, PBCR, and QACR in the coordinate plane.

Step 3: So the area of ∆QPR is calculated as 

Area of ∆PQR = [Area of trapezium PQAB + Area of trapezium PBCR] – [Area of trapezium QACR]      . . . (1)

Step 4: Now calculating areas of all 3 trapeziums.

Since Area of a trapezium = (1 / 2) (sum of the parallel sides) × (distance between sides)

Finding Area of a Trapezium PQAB  

⇒ Area of trapezium PQAB = (1 / 2)(QA + PB) × AB

⇒ QA = y₂

⇒ PB = y₁

⇒ AB = OB – OA = x₁ – x₂

⇒ Area of trapezium PQAB = (1 / 2)(y₁ + y₂)(x₁ – x₂ ) . . . (2)

Finding Area of a Trapezium PBCR

⇒ Area of trapezium PBCR =(1 / 2) (PB + CR) × BC

⇒ PB = y₁

⇒ CR = y₃

⇒ BC = OC – OB = x₃ – x₁

⇒ Area of trapezium PBCR =(1 / 2) (y₁ + y₃ )(x₃ – x₁) . . . (3)

Finding Area of a Trapezium QACR

⇒ Area of trapezium QACR = (1 / 2) (QA + CR) × AC

⇒ QA = y₂

⇒ CR = y₃

⇒ AC = OC – OA = x₃ – x₂

⇒ Area of trapezium QACR =(1 / 2)(y₂ + y₃ ) (x₃ – x₂ ). . . (4)

Step 5: Substituting (2), (3) and (4) in (1),

⇒ Area of ∆PQR = (1 / 2)[(y₁ + y₂)(x₁ – x₂ ) + (y₁ + y₃ )(x₃ – x₁) – (y₂ + y₃ ) (x₃ – x₂ )]

⇒ Area of ∆PQR = (1 / 2) |[x₁ (y₂ – y₃ ) + x₂ (y₃ – y₁ ) + x₃(y₁ – y₂)]|

Therefore, this is the formula to find the area of triangle if coordinates are given. 

Note: Observe that there is a mod, which indicates that, if we got a negative value we should only consider the numerical value as the area can’t be negative.

Example 1: What is the area of the ∆ABC whose vertices are A(1, 2), B(4, 2), and C(3, 5)?

Solution:

Firstly, let’s draw a diagram for a better understanding.

Now comparing the given coordinates with (x₁, y₁), (x₂, y₂), and (x₃, y₃).

Let, (x₁, y₁) = (1, 2)

⇒ (x₂, y₂) = (4, 2)

⇒ (x₃, y₃) = (3, 5)

Now we have to substitute the values in (1 / 2) |[x₁ (y₂ – y₃ ) + x₂ (y₃ – y₁ ) + x₃(y₁ – y₂)]|

⇒ (1 / 2) [1 (2 – 5 ) + 4 (5 – 2 ) + 3(2 – 2)]

⇒ (1 / 2) [(- 3) + 12 + 0]

⇒ (1 / 2) [9] = 4.5

Hence the area of the triangle is 4.5 sq units

Example 2: What is the value of x1 whose area of a triangle is 1 of coordinates (x₁, 1), (2, 3), and (4, 5)?

Solution:

Firstly, let’s draw a diagram for a better understanding.

In this problem, we have to find the value of ‘x1’, which is X coordinate of point A.

It is given that the area of the triangle is 1.

Now comparing the given coordinates with(x₁, y₁), (x₂, y₂), and (x₃, y₃).

Let, (x₁, y₁) = (x₁, 1)

⇒ (x₂, y₂) = (2, 3)

⇒ (x₃, y₃) = (4, 5)

Now we have to substitute the values in (1 / 2) |[x₁ (y₂ – y₃ ) + x₂ (y₃ – y₁ ) + x₃(y₁ – y₂)]|

⇒ (1 / 2) [x₁(3 – 5 ) + 2(5 – 1 ) + 4(1 – 3)] = 1

⇒ (1 / 2) [x₁(- 2) + 8 + -8] = 1

⇒ -x₁ = 1

⇒ x₁ = -1

Hence, the value of x₁ is -1.

There are other methods for finding area of triangle such as:

• Using Base and Height
• Using Heron’s Formula

Let’s discuss these methods as well.

Area of Triangle Using Base and Height

When the base and altitude of the triangle are given we will use this method and this method is the simplest of all the methods.

For a given triangle if the altitude of a triangle is ‘h‘ and the base of the triangle is ‘b‘ then the area of a triangle is given as:

Area of Triangle Using Heron’s Formula

When both the base and height of the triangle are not given then we can use Heron’s formula if the sides of the triangle are given.

If a, b, c are sides of the triangle then the area of the triangle is given as:

Read More,

• Coordinate Geometry
• Area of Triangle
• Area of Quadrilateral
• Area of Pentagon

Example 1: Find the area of the triangle whose height and base are 6 cm and 5 cm?

Solution: In the question, it is clearly mentioned that height and base are:

Given, h = 6 and b = 5

Area of triangle is given as =  (1 / 2) × b × h

⇒ (1 / 2) × 6 × 5

⇒ 3 × 5 = 15

Hence, the area of the given triangle is 15 cm²

Example 2: Find the height of the triangle whose area is 12 cm² and the base is 6 cm?

Solution: 

Given, area = 12 and b = 6

Area of triangle is = (1 / 2) × b × h

⇒ 12 = (1 / 2) × 6 × h

⇒ h = 12 / 3 = 4

Hence, the height of the given triangle is 4 cm.

Example 3: If the sides of the triangle are 3 cm, 4 cm, and 5 cm then find the area of the triangle.

Solution:

Let a = 3, b = 4, and  c = 5 

First, we have to find semi perimeter 

⇒ s = (a + b + c) / 2

⇒ s = (3 + 4 + 5) / 2 

⇒ s = 12 / 2 = 6

As we know heron’s formula is √[s(s – a)(s – b)(s – c)], so substituting values in it

⇒ √[s(s – a)(s – b)(s – c)]

⇒ √[6(6 – 3)(6 – 4)(6 – 5)]

⇒ √[6 × 3 × 2 × 1]

⇒ √36 = 6

The area of the triangle is 6 cm²

Example 4: Using heron’s formula derive formula to find area of equilateral triangle whose side is a

Solution:

To find semi perimeter

⇒ s = (a + b + c) / 2

⇒ s = (a + a + a ) / 2

⇒ s = 3a / 2

Now using heron’s formula

⇒ √[s(s – a)(s – b)(s – c)]

⇒ √[(3a / 2)((3a / 2) – a)((3a / 2) – a)((3a / 2) – a)]

⇒ √[(3a / 2)(a / 2)(a / 2)(a / 2)]

⇒ √(3a⁴ / 16)

⇒ √3(a²) / 4 

Hence area of a equilateral triangle is  √3(a²) / 4

What is the formula for area of triangle in coordinate geometry?

If coordinates of three vertices of triangle is given then formula for area of triangle is given as:

(1 / 2) |[x₁ (y₂ – y₃ ) + x₂ (y₃ – y₁ ) + x₃(y₁ – y₂)]|

What are some other formulas for area of triangle?

Some other formula for area of triangle include:

• A = 1/2 ​× base × height
• A = √[s(s−a)(s−b)(s−c)​] [Heron’s Formula]

Are there any alternate methods to find the area of a triangle in coordinate geometry?

Yes some other methods include:

• Using the distance formula to calculate the lengths of the sides and then applying Heron’s formula.
• Using vectors to determine the cross product, which can also provide the area of the triangle formed by two vectors originating from the same vertex.

What happens if the points are collinear?

If the points are collinear, they lie on a single straight line, and the area of the triangle they form is zero.