Question

A square on a coordinate plane has one vertex at (-0.5, -2) and a perimeter of 10 units. If all of the vertices are located in Quadrant III, what are the coordinates of the other three vertices.

Answer 1

Answer:

B(-3,-2), C(-3,-4.5),D(-0.5,-4.5)

Step-by-step explanation:

step 1

Find the length side of the square

we know that the perimeter of square is

$P=4b$

where

b is the length side of the square

we have

$P=10\ units$

substitute and solve for b

$10=4b$

Divide by 4 both sides

$b=2.5\ units$

step 2  

Find out the coordinates of the other three vertices

Let

A(-0.5,-2) ---> the given coordinates of one vertex

we know that

all of the vertices are located in Quadrant III

so

the other three vertices are located at the left and down of vertex A

coordinate of vertex B located at 2.5 units at left of vertex A

coordinate of vertex C located at 2.5 units at left and 2.5 units down of vertex A

coordinate of vertex D located at 2.5 units down of vertex A

therefore

B(-0.5-2.5,-2) -----> B(-3,-2)

C(-0.5-2.5,-2-2.5) -----> C(-3,-4.5)

D(-0.5,-2-2.5) -----> D(-0.5,-4.5)

see the attached figure to better understand the problem