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?
What you are looking for is called the shoelace formula:

\begin{align*}
\text{Area}
&amp;= \frac12 \big| (x_A - x_C) (y_B - y_A) - (x_A - x_B) (y_C - y_A) \big|\\
&amp;= \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|\\
&amp;= \frac12 \Big|\det \begin{bmatrix}
x_A &amp; x_B &amp; x_C \\
y_A &amp; y_B &amp; y_C \\
1 &amp; 1 &amp; 1
\end{bmatrix}\Big|
\end{align*}

The last version tells you 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 $n$ vertexes $(x_i,y_i)$ is

$$ \frac12\sum_{i=0}^{n-1} x_i y_{i+1} - x_{i+1} y_i $$

where indexes are added modulo $n$.
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
                 $$A = \frac{|-ad + b(a - c)|}{2}$$