Answer:
B) Isosceles
Step-by-step explanation:
The given triangle has vertices at A(-1,2), B(4,2) and C(3,-1).
We must first determine the length of the sides of the triangle, before we can classify it.
We apply the distance formula to find length of the sides.
[tex]|AB|=\sqrt{(4--1)^2+(2-2)^2}[/tex]
[tex]\Rightarrow |AB|=\sqrt{(4+1)^2+(2-2)^2}[/tex]
[tex]\Rightarrow |AB|=\sqrt{5^2+(0)^2}[/tex]
[tex]\Rightarrow |AB|=\sqrt{25}[/tex]
[tex]\Rightarrow |AB|=5 units[/tex].
The length of side BC
[tex]|BC|=\sqrt{(3-4)^2+(-1-2)^2}[/tex]
[tex]\Rightarrow |BC|=\sqrt{(-1)^2+(-3)^2}[/tex]
[tex]\Rightarrow |BC|=\sqrt{1+9}[/tex]
[tex]\Rightarrow |BC|=\sqrt{10}[/tex]
The length of side AC
[tex]|AC|=\sqrt{(3--1)^2+(-1-2)^2}[/tex]
We simplify to obtain;
[tex]|AC|=\sqrt{(3+1)^2+(-3)^2}[/tex]
[tex]\Rightarrow |AC|=\sqrt{(4)^2+(-3)^2}[/tex]
[tex]|AC|=\sqrt{16+9}[/tex]
[tex]|AC|=\sqrt{25}[/tex]
[tex]|AC|=5\:units[/tex]
Since [tex]|AC|=5\:units=|AB|[/tex], the given triangle is an isosceles triangle.
The correct answer is
