I f the vertices o a triangle are :
A (ax, ay), B (bx, by), C (cx, cy)
Then area of ABC is
$$Area\ =\ \frac{\left(ax\left(by\ -\ cy\right)+\ bx\ \left(cy\ -\ ay\right)+\ cx\ \left(ay\ -\ by\right)\right)}{2}$$
Given that ;
L(1,3), M(5,1), N(3,5)
$$Area\ =\ \frac{\left(1\left(1-5\right)+5\left(5-3\right)+3\left(3-1\right)\right)}{2}$$
$$Area\ =\ \frac{\left(-4\ +\ \left(25-15\right)+\left(9-3\right)\right)}{2}$$
$$Area\ =\ \frac{\left(-4+10+6\right)}{2}$$
$$Area\ =\ \frac{\left(-4+16\right)}{2}$$
$$Area\ =\ \frac{12}{2}$$
Area = 6

Option D is CORRECT.

AAPL wrote:What is the area of a triangle with the following vertices L(1, 3), M(5, 1), and N(3, 5)?

A. 3
B. 4
C. 5
D. 6
E. 7

I suggest subtracting from the area of a 4x4 square the sum of three right triangles (Si, Sii and Siii):
\[? = 16 - \left( {{S_{\,{\text{I}}}} + {S_{\,{\text{II}}}} + {S_{\,{\text{III}}}}} \right)\]
\[{S_{\,{\text{I}}}}\,\, = \,\,\frac{{2 \cdot 2}}{2} = 2\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,{S_{\,{\text{II}}}} = \,\,\frac{{2 \cdot 4}}{2} = {S_{\,{\text{III}}}} = 4\,\]
\[? = 16 - 10 = 6\]
(See figure attached.)

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.