we have that
\(Q (0, 5)\\R (-5, 0)\\S (-3, 4)\\T (-2, 3)\)
using a graph tool
see the attached figure
the Area of triangle QRS is equal to
\(A=\frac{1}{2}*b*h\\ A=\frac{1}{2}*RQ*ST\)
1) Find the distance RQ
Applying the formula of distance
\(d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}\)
\(dRQ=\sqrt{(0-5)^{2}+(-5-0)^{2}}\)
\(dRQ=\sqrt{(25)+(25)}\)
\(dRQ=\sqrt{50} units\)
2) Find the distance ST
\(dST=\sqrt{(3-4)^{2}+(-2+3)^{2}}\)
\(dST=\sqrt{(1)+(1)}\)
\(dST=\sqrt{2} units\)
3) Find the area of triangle QRS
\(A=\frac{1}{2}*RQ*ST\\\\ A=\frac{1}{2}*\sqrt{50}*\sqrt{2} \\\\ A=\frac{1}{2}*\sqrt{100} \\ \\ A=\frac{10}{2} \\ \\ A=5 units^{2}\)
therefore
the answer is the option
D- \(5\) square units
