# Finding out the area of a triangle if the coordinates of the three vertices are given

What is the simplest way to find out the area of a triangle if the coordinates of the three vertices are given in $x$-$y$ plane? One approach is to find the length of each side from the coordinates given and then apply Heron's formula. Is this the best way possible? Is it possible to compare the area of triangles with their coordinates provided without actually calculating side lengths?

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## 6 Answers

What you are looking for is the shoelace formula:

\begin{align*} \text{Area} &= \frac12 \big| (x_A - x_C) (y_B - y_A) - (x_A - x_B) (y_C - y_A) \big|\\ &= \frac12 \big| x_A y_B + x_B y_C + x_C y_A - x_A y_C - x_C y_B - x_B y_A \big|. \end{align*}

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Heron's formula is inefficient; there is in fact a direct formula. If the triangle has one vertex at the origin, and the other two vertices are $(a,b)$ and $(c,d)$, the formula for its area is $$A = \frac{\left| ad - bc \right|}{2}$$

To get a formula where the vertices can be anywhere, just subtract the coordinates of the third vertex from the coordinates of the other two (translating the triangle) and then use the above formula.

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if $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the vertices of a triangle then its area is given by

$$|\frac12(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))|$$

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The area $A$ of the triangle two of whose vertices lie on the axes, with coordinates $(a, 0)$, $(0, b)$, and a third vertex $(c, d)$ is obtained from previous formula by a mere horizontal axis shift of -a units as $$A = \frac{|-ad + b(a - c)|}{2}$$

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What "previous formula" do you mean? –  Rory Daulton May 25 at 11:41
Sorry to be vague about that ! I meant the formula A = |ad - bc|/2 for triangle with coordinates (a, b) , (c, d) and origin. –  Beedassy Lekraj May 27 at 4:56

The simplest way to remember how to calculate is by taking $\frac{1}{2}$ the value of the determinant of the matrix $$\begin{bmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{bmatrix}$$

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You know that AB x AC is a vector perpendicular to the plan ABC such that |AB x AC|= Area of the parallelogram ABA’C. Thus this area is equal to ½ |AB x AC|.

From AB= $(x_2 -x_1, y_2-y_1)$; AC= $(x_3-x_1, y_3-y_1)$, we deduce then

Area of $\Delta ABC$ = $\frac12$$[(x_2-x_1)(y_3-y_1)- (x_3-x_1)(y_2-y_1)]$

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