All points in Quadrant II have negative \(\displaystyle x\)-coordinates and positive \(\displaystyle y\)-coordinates. The only answer that fulfills these criteria is \(\displaystyle (-4,5)\). 

By definition, any point in the coordinate plane that has a non-zero value \(\displaystyle x\)-coordinate and a zero-value \(\displaystyle y\)-coordinate is on the \(\displaystyle x\)-axis.

A point with a positive \(\displaystyle x\)-coordinate and a negative \(\displaystyle y\)-coordinate can be found in Quadrant IV.

The \(\displaystyle x\)-coordinate of the midpoint can be found by dividing the sum of the \(\displaystyle x\)-coordinates of the endpoints by 2:

\(\displaystyle \left [(-10 ) + (-20) \right ] \div 2 = \left [-(10 + 20) \right ] \div 2 = -30 \div 2 = -15\)

The \(\displaystyle y\)-coordinate of the midpoint is found similarly:

\(\displaystyle \left [ (-20)+ 10 \right ] \div 2 = \left [-(20-10) \right ] \div 2 = -10 \div 2 = -5\)

The coordinates of the midpoint are \(\displaystyle (-15, -5)\).

The \(\displaystyle x\)-coordinate of the midpoint is half the sum of the \(\displaystyle x\)-coordinates of the endpoints:

\(\displaystyle \frac{1}{2} \left ( 4 \frac{1}{3} + 12 \right )\)

\(\displaystyle = \frac{1}{2} \times 16 \frac{1}{3}\)

\(\displaystyle = \frac{1}{2} \times \frac{49}{3}\)

\(\displaystyle = \frac{49}{6}\)

\(\displaystyle =8 \frac{1}{6}\)

The \(\displaystyle y\)-coordinate of the midpoint is found similarly:

\(\displaystyle \frac{1}{2} \left (- 4 \frac{1}{3} + 12 \right )\)

\(\displaystyle = \frac{1}{2} \left (12 - 4 \frac{1}{3} \right )\)

\(\displaystyle = \frac{1}{2} \times 7 \frac{2}{3}\)

\(\displaystyle = \frac{1}{2} \times \frac{23}{3}\)

\(\displaystyle = \frac{23}{6}\)

\(\displaystyle = 3 \frac{5}{6}\)

The midpoint has coordinates \(\displaystyle \left ( 8 \frac{1}{6}, 3 \frac{5}{6} \right )\).

By definition, a point with a negative \(\displaystyle x\)-coordinate and a negative \(\displaystyle y\)-coordinate lies in Quadrant III on the coordinate plane.

To find the fraction of the entire squares area that lays in quadrant II, notice that the square is taking up an equivalent amount in each of the four quadrants. Thus, \(\displaystyle \frac{1}{4}\) of the squares area is in quadrant II. 


Also note that the area of the square that lays in quadrant II is \(\displaystyle 25\) square units, thus the problem could have alternatively been solved by reducing \(\displaystyle \frac{25}{100}=\frac{1}{4}\).  

In order to solve this problem, apply the formula \(\displaystyle A=S^2\), in order to conclude that the length of side \(\displaystyle S\) must equal \(\displaystyle 7\) for the area to equal \(\displaystyle 49\). 

Once you've found the length of side \(\displaystyle S\), apply the formula \(\displaystyle P=4S\).

Thus, the solution is \(\displaystyle P=4\times7=28\)

In order to solve this problem, apply the formula \(\displaystyle A=S^2\), in order to conclude that the length of side \(\displaystyle S\) must equal \(\displaystyle 8\) for the area to equal \(\displaystyle 64\). 

Once you've found the length of side \(\displaystyle S\), apply the formula \(\displaystyle P=4S\)


\(\displaystyle P=4\times8=32\)

To find the perimeter of this square, apply the formula: \(\displaystyle P=4S\), where \(\displaystyle S=\) the length of one side of the square. 

Thus, the solution is:

\(\displaystyle P=4S\)
\(\displaystyle P=4\times6=24\)