## How to find area of triangle given three vertices?

We can find Area of triangle using formula ** base height** if we know the length of base and height of triangle.

We can also find area of triangle using **Heron's formula** if we know the length of three sides of triangle.

But, how can we find area of triangle if we know only the coordinates of vertices of triangle. If, we know the vertices of triangle then we can definitely use distance formula to find the length of all the sides which can enable us to use **Heron's formula** to find **area of triangle**. But, this will become too much lengthy and tedious.

**We have a formula which can be directly used on the vertices of triangle to find its area.**

**If, (x1, x2), (x2, y2) and (x3, y3) are the coordinates of vertices of triangle then**

**Area of Triangle = **

Now, we can easily derive this formula using a small diagram shown below.

Suppose, we have a as shown in the diagram and we want to find its area.

Let the coordinates of vertices areĀ **(x1, y1), (x2, y2) and (x3, y3).**

We draw perpendiculars AP, BQ and CR to x-axis.

**Area of = Area of Trapezium ABQP + Area of Trapezium BCRQ - Area of Trapezium ACRP**

Area of

**Example:** Find area of triangle whose vertices are (1, 1), (2, 3) and (4, 5)

**Solution:**

We have (x1, y1) = (1, 1), (x2, y2) = (2, 3) and (x3, y3) = (4, 5)

**Using formula:**

**Area of Triangle =**

**Because, Area cannot be negative. We only consider the numerical value of answer. Therefore, area of triangle = 1 sq units.**

This formula only works for the 1st quadrant of the coordinate system.

@Arif, this formula works for all 4 quadrants. Take one example and check. You just need to ignore negative sign at the end.

Thanks

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