What is the perimeter of a triangle with vertices located at (-1, 4), (2, 7), and (1, 5), rounded to the nearest hundredth?

You can find the perimeter by finding the distance between all three points and then adding the sums together.

From (-1, 4) to (2, 7)
d=√(x₂-x₁)²+(y₂-y₁)²
d=√(2-(-1)²+(7-4)²
d=√(3)²+(3)²
d=√9+9
d=√81
d=9

From (2, 7) to (1, 5)
d=√(x₂-x₁)²+(y₂-y₁)²
d=√(1-2)²+(5-7)²
d=√(-1)²+(-2)²
d=√1+4
d=√5
d=2.24

From (1, 5) to (-1, 4)
d=√(x₂-x₁)²+(y₂-y₁)²
d=√(-1-1)²+(4-5)²
d=√(-2)²+(-1)²
d=√4+1
d=√5
d=2.24

2.24+2.24+9
=13.48

The perimeter of the triangle is 13.48.

A. -10 is the value of x