IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry - not much difficult [admin]: finding area of triangle given the coordinates of vertices

edison

Below i provide the derivation

I make use of linear combination of areas of trapezoids (area of a trapezoid = (1/2)times "its height" times "the sum of the lengths of its parallel sides") and produces a simple algebraic expression for the area of a triangle. For the development of this result we start with triangle PQR in the first quadrant. (The location is for convenience and the final result is independent of the location of the triangle.) In Figure 1 we draw perpendicular segments from each vertex to the horizontal axis.

Figure 1.

We focus on the three trapezoids APQB, BQRC, and APRC and it follows that

area(

) = area(APQB) + area(BQRC) - area(APRC)

Computing the areas of trapezoids APQB, BQRC, and APRC we have

Substituting these results into the expression for area(

) and collecting terms we have

To make this expression independent of the order in which the vertices of the triangle are labeled and the location of the triangle we take the absolute value of the right side of the preceding expression and get

for any triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃). This expression for the area of a triangle requires no intermediate calculations such as the length of sides that is needed by Hero's formula.

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