Given a square with two vertices of one side located at -5,-3 and -5,12 in square units what is its area

Step-by-step explanation:

Lets find the distance (the side length of the square)

D=sqrt((x2-x1)^(2)+(y2-y1)^(2))

plug in the coordinates:

D=sqrt((-5-(-5))^(2)+(12-(-3))^(2))

D=sqrt(0^(2)+15^(2))

D=sqrt(15^(2))

D=15

Area of a square is side length squared

A=s^2

A=15^2

A=225 units squared

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step-by-step explanation:

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it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.