## INTRODUCTION:-

It is the second blog of coordinate geometry. In its previous blog, we have to discuss how to find the distance between two points in the cartesian plane by distance formula.

Now we will understand how to find the area of a triangle in the cartesian plane by the formula.

In some cases, we have coordinates of the triangle & in some cases, we need to find it by drawing a referencing line as we discuss in the previous blog.

For better understanding, you should read our blog COORDINATE GEOMETRY: How to find the distance between two points by Distance Formula?

## Terminology:-

*The cartesian plane, X-axis, Y-axis, Coordinate, Abscissa, Ordinate, Origin all terms explained in the previous blog with the picture.*

## How to find the area of a triangle in the Cartesian plane?

In the Cartesian plane first plot the given coordinates of the triangle.

*We have different ways of finding the area of a triangle-*

*By area of triangle formula (If right angle triangle).**By Heron's formula (But it is lengthy).**By three-point formula (Helpful in the cartesian plane).*

### Question

**Find the area of a triangle if coordinates of its vertices are (5,3), (3,1) & (1,4).**

### Answer

*We will use the three-point formula method to find the area of a triangle-*

*Area of a triangle = (1/2) {x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)}*

*First, we plot the coordinates of vertices of the triangle in the cartesian plane & then solve-*

*Here we have the following values of coordinates of x & y are -*

*x₁=5, y₁=3, x₂=3, y₂=1, x₃=1 & y₃=4**Put all the values in the above formula -*

*(1/2) {x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)}**(1/2) {5(1-4)+3(4-3)+1(3-1)}**(1/2){5×(-3)+(3×1)+(1×2)}**(1/2){-15+3+2}**(1/2){-10}**-10÷2**-5 sq. unit*

*Our area is -5 sq unit. But the area has a positive sign. So we take it 5 sq. unit.*

## CONCLUSION:-

We have an area of a triangle in the cartesian plane. So finally we learned how to find the area of a triangle in the cartesian plane by formula method. In the next blog, we will learn the section formula.