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 = 12(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))
Now, we can easily derive this formula using a small diagram shown below.
Suppose, we have a △ABC as shown in the diagram and we want to find its area.
Let the coordinates of vertices are (x1, y1), (x2, y2) and (x3, y3).
We draw perpendiculars AP, BQ and CR to x-axis.
Area of △ABC = Area of Trapezium ABQP + Area of Trapezium BCRQ - Area of Trapezium ACRP
⇒ Area of △ABC=12(y1+y2)(x1−x2)
+12(y1+y3)(x3−x1)
−12(y2+y3)(x3−x2)=12(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))
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Answers & Comments
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 = 12(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))
Now, we can easily derive this formula using a small diagram shown below.
Suppose, we have a △ABC as shown in the diagram and we want to find its area.
Let the coordinates of vertices are (x1, y1), (x2, y2) and (x3, y3).
We draw perpendiculars AP, BQ and CR to x-axis.
Area of △ABC = Area of Trapezium ABQP + Area of Trapezium BCRQ - Area of Trapezium ACRP
⇒ Area of △ABC=12(y1+y2)(x1−x2)
+12(y1+y3)(x3−x1)
−12(y2+y3)(x3−x2)=12(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))