Question 68245
I have plotted the lines for you. 
{{{ graph( 300, 200, 2, 12, -10, 10, (x/2)-6, 3-(x/3), 4x-16) }}} 


You can see where the 3 lines intercepet above.
These points can be found by simulataneously solving the 3 equations you worked out.
Let's do that now:
{{{x-2y=12}}}          ...(1)
{{{x+3y=9}}}           ...(2)
{{{4x-y=-16}}}         ...(3)

rearrange equation (1) to give x in terms of y
{{{x=2y+12}}} and put this new value for x into equation (2):
{{{2y+12+3y=9}}}
{{{y=-3/5}}}
If we put this value for y into the expression we had for x, we get:
{{{x=2(-3/5)+12}}}
{{{x=-54/5}}}

Following the process above similarly for equations (1) and (2), (2) and (3), finally (1) and (3) you will obtain the location of the three corners of the triangle in (x,y) coordinates.

Now find the lengths of the three sides by using:
{{{L=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}
where L is the length between the points (x1,y1) and (x2,y2). Do this for all three combinations of the 2 points you obtained earlier. Call these lengths a,b and c.

Now you have the lengths, you can calculate the 'semiperimeter', also called 's', using:
{{{s=(1/2)(a+b+c)}}}

Now we can use Heron's formula to find the area:
{{{area=sqrt(s(s-a)(s-b)(s-c))}}}

I hope this has helped you. If you have any further questions, please email me or visit my website at www.geocities.com/quibowibbler  .

Best Regards,
Adam