Answer :
The distance between two points A(x1, y2) and B(x2, y2) can be writenas,
[tex]AB=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2_{}}[/tex]The distance gives the length AB.
Hence,
[tex]\begin{gathered} AB=\sqrt[]{(-2-(-4)^{})^2+(3-1)^2}=\sqrt[]{2^2+2^2} \\ =\sqrt[]{8} \\ =2.83 \end{gathered}[/tex]The length BC is
[tex]\begin{gathered} BC=\sqrt[]{(3-(-2))^2+(-4-3)^2} \\ =\sqrt[]{5^2+(-7)^2}=\sqrt[]{25+49}=\sqrt[]{74} \\ =8.60 \end{gathered}[/tex]The length AC is
[tex]\begin{gathered} AC=\sqrt[]{(3-(-4))^2+(-4-1)^2}=\sqrt[]{7^2+(-5)^2} \\ =\sqrt[]{49+25}=\sqrt[]{74}=8.60 \end{gathered}[/tex]The perimeter of traingle ABC is,
AB+BC+AC=2.83+8.6+8.6=20.03
The perimeter of triangle ABC is 20.03.