Answer:
23.4 units
Step-by-step explanation:
Given the point of the vertices of triangle ABC as A(−5,1), B(2,1), and C(6,4).
The distance (d) between two points X (x₁ , y₁) and Y(x₂, y₂) is given as:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]. We can use this to calculate the length of the sides of the triangle. Therefore:
[tex]AB=\sqrt{(2-(-5))^2+(1-1)^2}=\sqrt{7^2}=7\\ AC=\sqrt{(6-(-5))^2+(4-1)^2}=\sqrt{11^2+3^2}=11.4\\\\BC=\sqrt{(6-(2))^2+(4-1)^2}=\sqrt{4^2+3^2}=5\\[/tex]
The perimeter of the triangle ABC = AB + AC + BC = 7 + 11.4 + 5 = 23.4 units
