Respuesta :
Use the next formula to find the distance between two points:
[tex]\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \end{gathered}[/tex]Sides of the triangle have the next lengths:
P1 to P2: (-4,-1) to (0,9)
[tex]\begin{gathered} P1P2=\sqrt{(0-(-4))^2+(9-(-1))^2} \\ \\ P1P2=\sqrt{4^2+10^2} \\ \\ P1P2=\sqrt{16+100} \\ \\ P1P2=\sqrt{116} \end{gathered}[/tex]P2 to P3: (0,9) to (3,2)
[tex]\begin{gathered} P2P3=\sqrt{(3-0)^2+(2-9)^2} \\ \\ P2P3=\sqrt{3^2+(-7)^2} \\ \\ P2P3=\sqrt{9+49} \\ \\ P2P3=\sqrt{58} \end{gathered}[/tex]P3 to P1: (3,2) to (-4,-1)
[tex]\begin{gathered} P3P1=\sqrt{(-4-3)^2+(-1-2)^2} \\ \\ P3P1=\sqrt{(-7)\placeholder{⬚}^2+(-3)\placeholder{⬚}^2} \\ \\ P3P1=\sqrt{49+9} \\ \\ P3P1=\sqrt{58} \end{gathered}[/tex]As the length of two sides of the triangle is the same it is a isosceles triangle.
To prove if it is a right triangle use the pythagorean theorem:
The greatest side needs to be the hypotenuse
[tex]\begin{gathered} hyp^2=Leg1^2+Leg2^2 \\ \\ (\sqrt{116})\placeholder{⬚}^2=(\sqrt{58})\placeholder{⬚}^2+(\sqrt{58})\placeholder{⬚}^2 \\ \\ 116=58+58 \\ \\ 116=116 \end{gathered}[/tex]The given triangle makes true the pythagorean theorem, it is a right triangle.
