Question

#8 is a triangle with vertices g(-5, 6), h(8, 5), and k(0, 10) a right triangle? it could be if angles are adjusted; not enough information; no; yes

Explanation:

Step1: Calculate side - lengths

Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For side $GH$, $x_1=-5,y_1 = 6,x_2 = 8,y_2=5$. Then $GH=\sqrt{(8 + 5)^2+(5 - 6)^2}=\sqrt{169 + 1}=\sqrt{170}$. For side $HK$, $x_1 = 8,y_1=5,x_2=0,y_2 = 10$. Then $HK=\sqrt{(0 - 8)^2+(10 - 5)^2}=\sqrt{64 + 25}=\sqrt{89}$. For side $KG$, $x_1=0,y_1 = 10,x_2=-5,y_2=6$. Then $KG=\sqrt{(-5 - 0)^2+(6 - 10)^2}=\sqrt{25+16}=\sqrt{41}$.

Step2: Check Pythagorean theorem

Check if $a^{2}+b^{2}=c^{2}$. $(\sqrt{41})^{2}+(\sqrt{89})^{2}=41 + 89=130
eq170=(\sqrt{170})^{2}$, $(\sqrt{41})^{2}+(\sqrt{170})^{2}
eq(\sqrt{89})^{2}$, $(\sqrt{89})^{2}+(\sqrt{170})^{2}
eq(\sqrt{41})^{2}$.

Answer:

No