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