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