# How can I find the area of a triangle given the coordinates of its corners ad (x1,y1),(x2,y2) and (x3,y3)?

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Let the verices of the triangle be A(x1,y1), B(x2,y2) and C(x3,y3) .

Then I imagine (I can not draw here) A , B ,C on a graph sheet.

AB, BC and CA form the trapeziums with respect to X axis.

Area of trapezium under AB with x axis = (Sum of the || sides)*(distance between the || lines)/2 = (y1+y2)(x2-x1)/2.

|||ly, we can say the area of the trapezium BC = (y2+y3)(x3-x2)/2. Area of the trapezium under AC with x axis = ((y3+y1)(x2-x10/2.

The bounded area by the triangle ABC = Area of trapeziums under AB with axis + BC with x axis - Area of trapezium under AC with x axis.

Area ABC = (1/2){(y2+y1)(x2-x1)+(y3+y2)(x3-x1)/2 - (y3+y1)(x3-x1)}

Area ABC = (1/2){(y1+y2)(x1-x2)+(y2+y3)(x3-x1)+(y3+1)((x1-x3)} in the cyclic form which is easy to remember. This you can also write in the form of a determinant.

Area of triangle =(1/2)| [(1 ,1,1 ), (x1 , x2 ,x3),(y1,y2,y3)] |

Given the coordinates of the three corners of a triangle, first take any two of them and consider them as the base. Let's take (x1, y1) and (x2, y2) to do the same here.

The distance between the two which is sqrt [(x1 - x2) ^2 + (y1 - y2) ^2] is the length of the base.

Now we need to find the height. Consider the line between the points that form the base: y-y1 = [(y2-y1) / (x2-x1)] *(x-x1)

A line perpendicular to this line has a slope - [(x2-x1) / (y2 - y1)] and it passes through the point (x3, y3). Therefore we get the equation of the perpendicular line. We can then find the point of intersection between the perpendicular and the base and determine the distance of (x3, y3) from the point of intersection. This is equal to the height.

The area of a triangle is (1/2)*b*h. As we know the base and the height we can find the area of the triangle.