Answer:
The answer is the option A
[tex]\frac{1}{2}(\sqrt{[(y2)^{2}+(x2)^{2}]*[(y1)^{2}+(x1)^{2}]})[/tex]
Step-by-step explanation:
see the attached figure with letters to better understand the problem
we know that
The area of the triangle is equal to
[tex]A=\frac{1}{2}bh[/tex]
where
b is the base
h is the height
In this problem
[tex]b=AB, h=AC[/tex]
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
Find the distance AB
[tex]A(0,0), B(x2,y2)[/tex]
substitute
[tex]dAB=\sqrt{(y2-0)^{2}+(x2-0)^{2}}[/tex]
[tex]dAB=\sqrt{(y2)^{2}+(x2)^{2}}[/tex]
Find the distance AC
[tex]A(0,0), C(x1,y1)[/tex]
substitute
[tex]dAC=\sqrt{(y1-0)^{2}+(x1-0)^{2}}[/tex]
[tex]dAC=\sqrt{(y1)^{2}+(x1)^{2}}[/tex]
Find the area of the triangle
we have
[tex]b=\sqrt{(y2)^{2}+(x2)^{2}}, h=\sqrt{(y1)^{2}+(x1)^{2}}[/tex]
substitute
[tex]A=\frac{1}{2}(\sqrt{(y2)^{2}+(x2)^{2}})(\sqrt{(y1)^{2}+(x1)^{2}})[/tex]
[tex]A=\frac{1}{2}(\sqrt{[(y2)^{2}+(x2)^{2}]*[(y1)^{2}+(x1)^{2}]})[/tex]
