# Area of Quadrilateral in Coordinate Geometry – Formulas & Examples

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A quadrilateral is a closed shape and a type of polygon that has four sides, four vertices, and four angles. It is formed by joining four non-collinear points. The area of a quadrilateral is the amount of region that is present inside it. We can find the area of regular quadrilaterals using formulas.

In this article, you will learn how to find the area of an irregular quadrilateral (or any type of quadrilateral) using coordinate geometry.

## Area of Quadrilateral in Coordinate Geometry

When a diagonal is drawn in a quadrilateral it is divided into two triangles. The area of the quadrilateral will be equal to the sum of the areas of these two triangles. To find the area of a quadrilateral, we find the area of individual triangles and add their areas.

In the above figure, quadrilateral $\text{ABCD}$ is divided into two triangles $\text{ABC}$ and $\text{CDA}$ by a diagonal $\text{AC}$.

Area of $\text{Quad. ABCD} = \text{Area of } \triangle \text{ABC} + \text{ Area of } \triangle \text{CDA}$.

In the above figure, quadrilateral $\text{ABCD}$ is divided into two triangles $\text{ABD}$ and $\text{BCD}$ by a diagonal $\text{BD}$.

Area of $\text{Quad. ABCD} = \text{Area of } \triangle \text{ABD} + \text{ Area of } \triangle \text{BCD}$.

When the coordinates of vertices of a quadrilateral are known then we can find the area of a quadrilateral by calculating the areas of two triangles

In the above figure,

Area of $\triangle \text{ABC} = |\frac{1}{2}[x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)]|$

Area of $\triangle \text{ACD} = |\frac{1}{2}[x_1(y_3 – y_4) + x_3(y_4 – y_1) + x_4(y_1 – y_3)]|$

Therefore area of quadrilateral $\text{ABCD} = |\frac{1}{2}[x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)]| + |\frac{1}{2}[x_1(y_3 – y_4) + x_3(y_4 – y_1) + x_4(y_1 – y_3)]|$

$= \frac{1}{2}|[x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)] + [x_1(y_3 – y_4) + x_3(y_4 – y_1) + x_4(y_1 – y_3)]|$

$= \frac{1}{2}|[x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) + x_1(y_3 – y_4) + x_3(y_4 – y_1) + x_4(y_1 – y_3)]|$

$= \frac{1}{2}|[x_1(y_2 – y_3) + x_1(y_3 – y_4) + x_2(y_3 – y_1) + x_3(y_1 – y_2) + x_3(y_4 – y_1) + x_4(y_1 – y_3)]|$

$= \frac{1}{2}|[x_1(y_2 – y_3 + y_3 – y_4) + x_2(y_3 – y_1) + x_3(y_1 – y_2 + y_4 – y_1) + x_4(y_1 – y_3)]|$

$= \frac{1}{2}|[x_1(y_2 – y_4) + x_2(y_3 – y_1) + x_3(- y_2 + y_4) + x_4(y_1 – y_3)]|$

$= \frac{1}{2}|[x_1(y_2 – y_4) + x_2(y_3 – y_1) + x_3(y_4 – y_2) + x_4(y_1 – y_3)]|$

$= \frac{1}{2}|x_1 y_2 – x_1 y_4 + x_2 y_3 – x_2 y_1 + x_3 y_4 – x_3 y_2 + x_4 y_1 – x_4 y_3|$

$= \frac{1}{2}|x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1 – x_1 y_4 – x_2 y_1 – x_3 y_2 – x_4 y_3|$

$= \frac{1}{2}|(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1) – (x_2 y_1 + x_3 y_2 + x_4 y_3 + x_1 y_4)|$

$= \frac{1}{2}|(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1) – (y_1 x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1)|$

In the above figure,

Area of $\triangle \text{ABD} = |\frac{1}{2}[x_1(y_2 – y_4) + x_2(y_4 – y_1) + x_4(y_1 – y_2)]|$

Area of $\triangle \text{BCD} = |\frac{1}{2}[x_2(y_3 – y_4) + x_3(y_4 – y_2) + x_4(y_2 – y_3)]|$

Therefore area of quadrilateral $\text{ABCD} = |\frac{1}{2}[x_1(y_2 – y_4) + x_2(y_4 – y_1) + x_4(y_1 – y_2)]| + |\frac{1}{2}[x_2(y_3 – y_4) + x_3(y_4 – y_2) + x_4(y_2 – y_3)]|$

$= \frac{1}{2}|[x_1(y_2 – y_4) + x_2(y_4 – y_1) + x_4(y_1 – y_2)] + [x_2(y_3 – y_4) + x_3(y_4 – y_2) + x_4(y_2 – y_3)]|$

$= \frac{1}{2}|[x_1(y_2 – y_4) + x_2(y_4 – y_1) + x_4(y_1 – y_2) + x_2(y_3 – y_4) + x_3(y_4 – y_2) + x_4(y_2 – y_3)]|$

$= \frac{1}{2}|[x_1(y_2 – y_4) + x_2(y_4 – y_1) + x_2(y_3 – y_4) + x_3(y_4 – y_2) + x_4(y_2 – y_3) + x_4(y_1 – y_2)]|$

$= \frac{1}{2}|[x_1(y_2 – y_4) + x_2(y_4 – y_1 + y_3 – y_4) + x_3(y_4 – y_2) + x_4(y_2 – y_3 + y_1 – y_2)]|$

$= \frac{1}{2}|[x_1(y_2 – y_4) + x_2(- y_1 + y_3) + x_3(y_4 – y_2) + x_4(- y_3 + y_1)]|$

$= \frac{1}{2}|x_1 y_2 – x_1 y_4 + x_2 y_3 – x_2 y_1 + x_3 y_4 – x_3 y_2 + x_4 y_1 – x_4 y_3|$

$= \frac{1}{2}|(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1) – (x_2 y_1 + x_3 y_2 + x_4 y_3 + x_1 y_4)|$

$= \frac{1}{2}|(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1) – (y_1x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1)|$

### Examples of Area of Quadrilateral in Coordinate Geometry

Example 1: Find area of a quadrilateral with vertices $\text{A}(−3, 1)$, $\text{B}(−1, 4)$, $\text{C}(3, 2)$, and $\text{D}(1, −2)$.

Area of quadrilateral with vertices $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, $\text{C}(x_3, y_3)$, and $\text{D}(x_4, y_4)$ is given by $\frac{1}{2}|(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1) – (y_1x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1)|$.

Therefore area of quadrilateral with vertices $\text{A}(−3, 1)$, $\text{B}(−1, 4)$, $\text{C}(3, 2)$, and $\text{D}(1, −2)$

$= \frac{1}{2}|(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1) – (y_1x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1)|$

$= \frac{1}{2}|(-3 \times 4 + (-1) \times 2 + 3 \times (-2) + 1 \times 1) – (1 \times (-1) + 4 \times 3 + 2 \times 1 + (-2) \times (-3))|$

$= \frac{1}{2}|(-12 – 2 – 6 + 1) – (-1 + 12 + 2 + 6)|$

$= \frac{1}{2}|-19 – 19|$

$= \frac{1}{2} \times |-38|$

$= \frac{1}{2} \times 38 = 19$ sq units

Example 2: Find area of a quadrilateral with vertices $\text{A}(-3, 2)$, $\text{B}(5, 4)$, $\text{C}(7, -6)$ and $\text{D}(-5, -4)$.

Area of quadrilateral with vertices $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, $\text{C}(x_3, y_3)$, and $\text{D}(x_4, y_4)$ is given by $\frac{1}{2}|(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1) – (y_1x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1)|$.

Therefore area of quadrilateral with vertices $\text{A}(-3, 2)$, $\text{B}(5, 4)$, $\text{C}(7, -6)$ and $\text{D}(-5, -4)$

$= \frac{1}{2}|(-3 \times 4 + 5 \times (-6) + 7 \times (-4) – 5 \times 2) – (2 \times 5 + 4 \times 7 + (-6) \times (-5) – 4 \times (-3))|$

$= \frac{1}{2}|(-12 – 30 – 28 – 10) – (10 + 28 + 30 + 12)|$

$= \frac{1}{2}|-80 – 80|$

$= \frac{1}{2} \times |-160|$

$= \frac{1}{2} \times 160 = 80$ sq units.

## Practice Problems

1. Find the area of the quadrilateral whose coordinates are $\text{A}(3,2)$, $\text{B}(5,4)$, $\text{C}(7,6)$, and $\text{D}(5,4)$, in that sequence.
2. Determine the area of the quadrilateral with the following vertices $(7, 5)$, $(4, 10)$, $(-6, 11)$, and $(-5,2)$.
3. Find the area of the quadrilaterals, the coordinates of whose vertices are $(1,2)$, $(6,2)$, $(5,3)$, and $(3,4)$.
4. Find the area of the quadrilateral, the coordinates of whose angular points are taken in order are $(1,1)$, $(3,4)$, $(5,−2)$, and $(4,−7)$.

## FAQs

### How do you find the area of a quadrilateral in coordinate geometry?

The area of a quadrilateral with vertices $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, $\text{C}(x_3, y_3)$, and $\text{D}(x_4, y_4)$ is given by the formula $\frac{1}{2}|(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1) – (y_1x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1)|$.

### What is the area of the quadrilateral?

The area of the quadrilateral is the region enclosed by the four sides of this polygon. Its unit of measurement is $\text{m}^2$.

### What is called a quadrilateral?

A quadrilateral is a plane figure that has four sides (or edges), and also has four corners (or vertices). The angles are present at the four vertices or corners of the quadrilateral. If $\text{ABCD}$ is a quadrilateral then angles at the vertices are $\angle \text{A}$, $\angle \text{B}$, $\angle \text{C}$ and $\angle \text{A}$. The sides of a quadrilateral are $\text{AB}$, $\text{BC}$, $\text{CD}$ and $\text{DA}$.

## Conclusion

The area of a quadrilateral with known coordinates of vertices is calculated by using the formula $\frac{1}{2}|(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1) – (y_1x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1)|$. This formula is an extension of the area of the triangle formula in coordinate geometry.