Question

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

Answer 1

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.

$|AB|=\sqrt{(4--1)^2+(2-2)^2}$

$\Rightarrow |AB|=\sqrt{(4+1)^2+(2-2)^2}$

$\Rightarrow |AB|=\sqrt{5^2+(0)^2}$

$\Rightarrow |AB|=\sqrt{25}$

$\Rightarrow |AB|=5 units$.

The length of side BC

$|BC|=\sqrt{(3-4)^2+(-1-2)^2}$

$\Rightarrow |BC|=\sqrt{(-1)^2+(-3)^2}$

$\Rightarrow |BC|=\sqrt{1+9}$

$\Rightarrow |BC|=\sqrt{10}$

The length of side AC

$|AC|=\sqrt{(3--1)^2+(-1-2)^2}$

We simplify to obtain;

$|AC|=\sqrt{(3+1)^2+(-3)^2}$

$\Rightarrow |AC|=\sqrt{(4)^2+(-3)^2}$

$|AC|=\sqrt{16+9}$

$|AC|=\sqrt{25}$

$|AC|=5\:units$

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

The correct answer is

Answer 2

Answer:

B) isosceles

Step-by-step explanation: