Prerequisites - P.6 - The Rectangular Coordinate System and Graph - Exercises - Page 59: 49

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Let's find the distance between the points using the distance formula, $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$So, $d_1=\sqrt{(-8-2)^2+(4-11)^2}=\sqrt{100+49}=\sqrt{149}$ $d_2=\sqrt{(-8+5)^2+(4-1)^2}=\sqrt{9+9}=\sqrt{18}$ $d_3=\sqrt{(2+5)^2+(11-1)^2}=\sqrt{49+100}=\sqrt{149}$ We see that two of the distances are equal, so the points form an isosceles triangle.