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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?
What you are looking for is called 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|\\ &= \frac12 \Big|\det \begin{bmatrix} x_A & x_B & x_C \\ y_A & y_B & y_C \\ 1 & 1 & 1 \end{bmatrix}\Big| \end{align*}
The last line indicates how to generalize the formula to higher dimensions.
PS. Another generalization of the formula is obtained by noting that it follows from a discrete version of the Green's theorem:
$$ \text{Area} = \iint_{\text{domain}}dx\,dy = \frac12\oint_{\text{boundary}}x\,dy - y\,dx $$
Thus the signed (oriented) area of a polygon with $n$ vertices $(x_i,y_i)$ is given by
$$ \text{Area} = \frac12\sum_{i=0}^{n-1} x_i y_{i+1} - x_{i+1} y_i $$
where indices are added modulo $n$.
You know that AB × AC is a vector perpendicular to the plane ABC such that |AB × AC|= Area of the parallelogram ABA’C. Thus this area is equal to ½ |AB × 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$$[(x_2-x_1)(y_3-y_1)- (x_3-x_1)(y_2-y_1)]$
Heron's formula is inefficient; there is in fact a direct formula. If the triangle has one vertex at the origin, and the other two vertices are $(a,b)$ and $(c,d)$, the formula for its area is $$ A = \frac{\left| ad - bc \right|}{2} $$
To get a formula where the vertices can be anywhere, just subtract the coordinates of the third vertex from the coordinates of the other two (translating the triangle) and then use the above formula.
The simplest way to remember how to calculate is by taking $\frac{1}{2}$ the value of the determinant of the matrix $$ \begin{bmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{bmatrix} $$
if $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the vertices of a triangle then its area is given by:-
$$\frac{ 1}{2}|(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))|$$
The area of a triangle $P(x_1, y_1)$, $Q(x_2, y_2)$ and $R(x_3, y_3)$ is given by $$\triangle= \left|\frac{1}{2}(x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2}))\right| $$ If the area of triangle is zero, it means the points are collinear.
If we code this in Python3, it will look like
def triangle_area(x1, y1, x2, y2, x3, y3):
return abs(0.5*(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2)))
If we code in js, it will look like this
function triangle_area(x1, y1, x2, y2, x3, y3){
return Math.abs(0.5*(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2)))
}
Now here's a solution which works in any vector space with an inner product: take the half of the root of the Gram-Determinant of two sides of the triangle, that is $$\frac12\sqrt{\det\begin{pmatrix}\langle b-a,b-a\rangle& \langle b-a,c-a\rangle\\ \langle b-a,c-a\rangle & \langle c-a,c-a\rangle \end{pmatrix}}.$$
For fun, I'll just throw out the really long way that I learned in 3rd grade, only because it hasn't been submitted. I don't endorse this, the Shoelace/Surveyor's formula is way better.
Two preparatory steps:
Then the area is then $F = \frac{1}{2} b v$.
The area $A$ of the triangle two of whose vertices lie on the axes, with coordinates $(a, 0)$, $(0, b)$, and a third vertex $(c, d)$ is obtained from previous formula by a mere horizontal axis shift of -a units as $$A = \frac{|-ad + b(a - c)|}{2}$$
