Question

state whether triangle ABC is scalene, isosceles or equilateral, A(1,-1), B(-4,6) and C(-3,-5) are the given coordinates.
Any help is appreciated.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{points}\mathrm{are}:\mathrm{A}\left(1,-1\right);\mathrm{B}\left(-4,6\right)\mathrm{and}\mathrm{C}\left(-3,-5\right). \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{distance}\mathrm{between}2\mathrm{points}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\mathrm{and}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)\mathrm{is}\mathrm{given}\mathrm{by} \\ \mathrm{distance}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2} \\ \mathrm{AB}=\sqrt{{\left(-4-1\right)}^2+{\left(6+1\right)}^2}=\sqrt{25+49}=\sqrt{74}\mathrm{units} \\ \mathrm{BC}=\sqrt{{\left(-3+4\right)}^2+{\left(-5-6\right)}^2}=\sqrt{1+121}=\sqrt{122}\mathrm{units} \\ \mathrm{CA}=\sqrt{{\left(-3-1\right)}^2+{\left(-5+1\right)}^2}=\sqrt{16+16}=\sqrt{32}\mathrm{units} \\ \mathrm{Now},\mathrm{in}∆\mathrm{ABC}\mathrm{each}\mathrm{side}\mathrm{are}\mathrm{of}\mathrm{unequal}\mathrm{measures},\mathrm{so} \\ ∆\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{scalene}∆. \end{gathered}$$