Question 174202
finding the area of triangle ABC 
whose vertices are
A(3,4)
B(7,3)
C(5,1)
<pre><font size = 4 color = "indigo"><b>
Let's draw the triangle:
{{{drawing(400,266.67,-1,8,-1,5,
graph(400,266.67,-1,8,-1,5),
triangle(3,4,7,3,5,1),
locate(2.8,4,A), locate(7,3,B),locate(5,1,C),
locate(6,2,a),locate(5,3.84,c),locate(3.9,2.5,b)
)}}}

{{{Area=sqrt(s(s-a)(s-b)(s-c))}}}
{{{s=(a+b+c)/2}}}

"s" stands for "semi-perimeter".

We find the length of each side using the distance
formula:

{{{matrix(1,3,d,"=",sqrt((x[2]-x[1])^2+(y[2]-y[1])^2))}}}

To find {{{a}}}, use the coordinates of its 
endpoints B and C: 

{{{matrix(1,3,a,"=",sqrt((7-5)^2+(3-1)^2))}}}
{{{matrix(1,3,a,"=",sqrt((2)^2+(2)^2))}}}
{{{matrix(1,3,a,"=",sqrt(4+4))}}}
{{{matrix(1,3,a,"=",sqrt(8))}}}
{{{matrix(1,3,a,"=",sqrt(4*2))}}}
{{{matrix(1,3,a,"=",sqrt(4)sqrt(2))}}}
{{{matrix(1,3,a,"=",2sqrt(2)=2.828427125)}}}

To find {{{b}}}, use the coordinates of its 
endpoints A and C: 

{{{matrix(1,3,b,"=",sqrt((5-3)^2+(1-4)^2))}}}
{{{matrix(1,3,b,"=",sqrt((2)^2+(-3)^2))}}}
{{{matrix(1,3,b,"=",sqrt(4+9))}}}
{{{matrix(1,3,b,"=",sqrt(13)=3.605551275)}}}

To find {{{c}}}, use the coordinates of its 
endpoints A and B: 

{{{matrix(1,3,c,"=",sqrt((7-3)^2+(3-4)^2))}}}
{{{matrix(1,3,c,"=",sqrt((4)^2+(-1)^2))}}}
{{{matrix(1,3,c,"=",sqrt(16+1))}}}
{{{matrix(1,3,c,"=",sqrt(17)=4.123105626)}}}

{{{semi-perimeter=s= (a+b+c)/2 = (2sqrt(2)+sqrt(13)+sqrt(17))/2=5.278542013}}}

{{{s-a=5.278542013-2.828427125=2.450114888}}}
{{{s-b=5.278542013-3.605551275=1.672990737}}}
{{{s-c=5.278542013-4.123105626=1.155436387}}}

{{{Area=sqrt(s(s-a)(s-b)(s-c))=sqrt(5.278542013(2.450114888)(1.672990737)(1.155436387))=sqrt(25)=5}}}

So the area = 5.

----------------------

However the easiest way to find the area is with a special
determinant formula:

{{{matrix(1,5,  Area, "=", 1/2, "×", abs(abs(matrix(3,3,3,4,1,7,3,1,5,1,1) )  )) }}}

where the first two elements in each row of the 
determinant are the coordinates of the three points,
and the last element in each row is 1.  That is, 
the last column contains all 1's.  Also notice that
we must take the absolute value of the determinant,
which is the reason for the double bars around the
determinant.

Evaluate the 3x3 determinant.  I assume you know 
how if you want to use this method.

{{{matrix(1,15,  Area, "=", 1/2, "×", abs(abs(matrix(3,3,3,4,1,7,3,1,5,1,1) )  ), "=", 1/2,"×",abs(-10),"=",1/2,"×",10, "=",5 )}}}

Edwin</pre>