The coordinates G(7,3), H(9, 0), (5, -1) form what type of polygon?an obtuse trianglean acute triangleO an equilateral trianglea right triangle

Answer:Is an acute triangleStep-by-step explanation:we know thatthe formula to calculate the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]we haveG(7,3), H(9, 0), I(5, -1) step 1Find the distance GHsubstitute in the formula[tex]d=\sqrt{(0-3)^{2}+(9-7)^{2}}[/tex][tex]d=\sqrt{(-3)^{2}+(2)^{2}}[/tex][tex]GH=\sqrt{13}\ units[/tex]step 2Find the distance IHsubstitute in the formula[tex]d=\sqrt{(0+1)^{2}+(9-5)^{2}}[/tex][tex]d=\sqrt{(1)^{2}+(4)^{2}}[/tex][tex]IH=\sqrt{17}\ units[/tex]step 3Find the distance GIsubstitute in the formula[tex]d=\sqrt{(-1-3)^{2}+(5-7)^{2}}[/tex][tex]d=\sqrt{(-4)^{2}+(-2)^{2}}[/tex][tex]GI=\sqrt{20}\ units[/tex]step 4Verify what type of triangle is the polygonwe know thatIf applying the Pythagoras Theorem[tex]c^{2}=a^{2}+b^{2}[/tex] ----> is a right triangle[tex]c^{2}> a^{2}+b^{2}[/tex] ----> is an obtuse triangle[tex]c^{2}< a^{2}+b^{2}[/tex] ----> is an acute trianglewhere c is the greater sidewe have[tex]c=\sqrt{20}\ units[/tex][tex]a=\sqrt{17}\ units[/tex][tex]b=\sqrt{13}\ units[/tex]substitute[tex]c^{2}= (\sqrt{20})^{2}=20[/tex][tex]a^{2}+b^{2}=(\sqrt{17})^{2}+(\sqrt{13})^{2}=30[/tex]therefore[tex]c^{2}< a^{2}+b^{2}[/tex]Is an acute triangle