Question

The area of a square I 41 square units. Two of its vertices are located at (0,-3) and (4,2). Select the point that could be a third vertex of this square.
A. (0,7)
B. (-5,1)
C.(5,13)
D.(-2,-2)
E.(4,-6)



There may be more than one answer by the way .​

Answer 1

The vertices of the square are the coordinates of the endpoints

The third vertex of the square could be (0,7) or (-5,1)

How to determine the third vertex

Two vertices of the square are:

(0,-3) and (4,2)

Calculate the distance between these vertices using:

$d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$

Square both sides

$d^2 = (x_2 - x_1)^2+(y_2 - y_1)^2$

The above expression represents the area of the square.

So, we have:

$(x_2 - x_1)^2+(y_2 - y_1)^2 = 41$

Next, we test the options with the given vertices (0,-3) and (4,2)

So, we have:

A. (0,7)

$(0 - 0)^2+(7 + 3)^2 = 41$

$100 \ne 41$

$(0 - 4)^2+(7 - 2)^2 = 41$

$41 = 41$

B. (-5,1)

$(-5 - 0)^2+(1 + 3)^2 = 41$

$41 = 41$

$(-5 - 4)^2+(1 -2)^2 = 41$

$82 \ne 41$

C.(5,13)

$(5 - 0)^2+(13 + 3)^2 = 41$

$281 \ne 41$

$(5 - 4)^2+(13 -2)^2 = 41$

$122 \ne 41$

D.(-2,-2)

$(-2 - 0)^2+(-2 + 3)^2 = 41$

$5 \ne 41$

$(-2 - 4)^2+(-2 -2)^2 = 41$

$52 \ne 41$

E.(4,-6)

$(4 - 0)^2+(-6 + 3)^2 = 41$

$25 \ne 41$

$(4 - 4)^2+(-6 -2)^2 = 41$

$64 \ne 41$

The vertices where at least one of the equations is true could be a third vertex of the square

Hence, a third vertex of the square could be (0,7) or (-5,1)

Read more about square vertices at:

brainly.com/question/1292795