Triangle ABC has vertices at A(−1 , 2), B(4, 2), and C(3, −1). Classify the triangle according to the side lengths. A) equilateral B) isosceles C) right D) scalene

Question

asked Feb 15, 2019 74.4k views

2 Answers

Justin Tamblyn · Answer 1 · 2019-02-20T12:27:38+0000

Answer:

B) Isosceles

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.

|AB|=√((4--1)^2+(2-2)^2)

$\Rightarrow |AB|=√((4+1)^2+(2-2)^2)$

$\Rightarrow |AB|=√(5^2+(0)^2)$

$\Rightarrow |AB|=√(25)$

$\Rightarrow |AB|=5 units$ .

The length of side BC

|BC|=√((3-4)^2+(-1-2)^2)

$\Rightarrow |BC|=√((-1)^2+(-3)^2)$

$\Rightarrow |BC|=√(1+9)$

$\Rightarrow |BC|=√(10)$

The length of side AC

|AC|=√((3--1)^2+(-1-2)^2)

We simplify to obtain;

|AC|=√((3+1)^2+(-3)^2)

$\Rightarrow |AC|=√((4)^2+(-3)^2)$

|AC|=√(16+9)

|AC|=√(25)

$|AC|=5\:units$

Since
$|AC|=5\:units=|AB|$ , the given triangle is an isosceles triangle.

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