How can I find the area of a triangle given the coordinates of its corners ad (x1,y1),(x2,y2) and (x3,y3)?

Let the verices of the triangle be  A(x1,y1), B(x2,y2) and
C(x3,y3) .

Then I imagine (I can not draw here) A , B ,C on
a graph sheet.

AB, BC and CA form the trapeziums with
respect to X axis.

Area of trapezium under AB with x axis
=  (Sum of the || sides)*(distance between the || lines)/2 =
(y1+y2)(x2-x1)/2.

|||ly, we can say the area of the
trapezium BC = (y2+y3)(x3-x2)/2. Area of the trapezium under AC with x axis =
((y3+y1)(x2-x10/2.

The   bounded area by the triangle ABC =
Area of  trapeziums under AB  with axis + BC with  x axis - Area of trapezium under AC
with x axis.

Area ABC =
(1/2){(y2+y1)(x2-x1)+(y3+y2)(x3-x1)/2 -
(y3+y1)(x3-x1)}

Area ABC =
(1/2){(y1+y2)(x1-x2)+(y2+y3)(x3-x1)+(y3+1)((x1-x3)} in the cyclic form which is easy to
remember. This you can also write in the form of a
determinant.

Area of triangle =(1/2)| [(1 ,1,1 ), (x1 , x2
,x3),(y1,y2,y3)] |