Use the slope formula, substituting \(\displaystyle x_{1}= 4, y_{1}=2,x_{2}= -8, y_{2}=-3\):

\(\displaystyle m =\frac{ y_{2}- y_{1} }{x_{2} - x_{1}}=\frac{ -3-2 }{-8-4} = \frac{ -5 }{-12} = \frac{5 }{12}\)

Using the slope formula, substituting \(\displaystyle x_{1}=5\), \(\displaystyle x_{2}=2\), \(\displaystyle y_{1}=5\), and \(\displaystyle y_{2}=4\):

\(\displaystyle m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\displaystyle m=\frac{4-5}{2-5}\)

Subtract to get:

\(\displaystyle m=\frac{-1}{-3}\)

Cancel out the negative signs to get:

\(\displaystyle m=\frac{1}{3}\)

Using the slope formula with \(\displaystyle x_{1}=-7\), \(\displaystyle x_{2}=2\), \(\displaystyle y_{1}=4\), \(\displaystyle y_{2}=5\):

\(\displaystyle m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\displaystyle m=\frac{5-4}{2-(-7)}\)

Subtract to get:

\(\displaystyle m=\frac{1}{9}\) 

Using the slope formula for 

\(\displaystyle x_{1}=-5\), \(\displaystyle x_{2}=1\), \(\displaystyle y_{1}=3\), and \(\displaystyle y_{2}=9\):

\(\displaystyle m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\displaystyle m=\frac{9-3}{1-(-5)}\)

Combine the negative signs to get:

\(\displaystyle m=\frac{9-3}{1+5}\)

Subtract and add to get:

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

Reduce to get:

\(\displaystyle m=1\)

Using the slope formula, where \(\displaystyle m\) is the slope, \(\displaystyle (x_{1},y_{1})=\) \(\displaystyle (4,7)\), and \(\displaystyle (x_{2},y_{2})=\) \(\displaystyle (2,3)\):

\(\displaystyle m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\displaystyle m=\frac{3-7}{2-4}\)

\(\displaystyle m=\frac{-4}{-2}\)

\(\displaystyle m=2\)

Using the slope formula, where \(\displaystyle m\) is the slope, \(\displaystyle (x_{1},y_{1})=\) \(\displaystyle (5,-4)\), and \(\displaystyle (x_{2},y_{2})=\) \(\displaystyle (-5,-4)\) :

\(\displaystyle m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\displaystyle m=\frac{-4-(-4)}{-5-5}\)

\(\displaystyle m=\frac{-4+4}{-10}\)

\(\displaystyle m=\frac{0}{-10}\)

\(\displaystyle m=0\)

Using the slope formula, where \(\displaystyle m\) is the slope, \(\displaystyle (x_{1},y_{1})=\) \(\displaystyle (-1,0)\), and \(\displaystyle (x_{2},y_{2})=\) \(\displaystyle (-10, 5)\) :

\(\displaystyle m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\displaystyle m=\frac{5-0}{-10-(-1)}\)

\(\displaystyle m=\frac{5}{-10+1}\)

\(\displaystyle m=-\frac{5}{9}\)

The value of the slope (m) is rise over run, and can be calculated with the formula below:

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

The coordinates of the first end of the plank would be (0,3), given that this is the starting point of the plank (so x would be 0), and y would be 3 since the brick is 3 inches tall. 

The coordinates of the second end of the plank would be (18,9) since the plank is 18 inches long (so x would be 18) and y would be 9 since the box was 9 inches tall at the other end. 

From this information we know that we can assign the following coordinates for the equation:

\(\displaystyle \left(x_{1},y_{1}\right) = (0,3)\) and \(\displaystyle (x_{2}, y_{2})= (18,9)\)

Below is the solution we would get from plugging this information into the equation for slope:

\(\displaystyle m = \frac{9-3}{18-0}}\)

This reduces to \(\displaystyle \frac{6}{18} =\frac{1}{3}\)

Because this is a right triangle, the area formula is simply:

 \(\displaystyle A = \frac{base \times height}{2}\)

Thus, the solution is:

\(\displaystyle A = \frac{10 \times 12}{2} = \frac{120}{2} = 60 \text{ square units}\)

Using the coordinate system, one can see the base of the triangle is 6 units in length. Since it is an equilateral triangle, the other two sides must also be 6 units each in length. Therefore the perimeter is:

\(\displaystyle P = a + b + c = 6 + 6 + 6 = 18\)