so the triangle has the vertices of (2, 18) (-2, -4), and (6,12), that gives us the endpoints for each line of
(2, 18) , (-2, -4)
(-2, -4) , (6,12)
(6,12) , (2, 18)

![\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-(-4)}{6-(-2)}\implies \cfrac{12+4}{6+2}\implies \cfrac{16}{8}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-4)=2[x-(-2)] \\\\\\ y+4=2(x+2)\implies y+4=2x+4\implies \blacktriangleright y=2x \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-4%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B12%7D%29%20%5C%5C%5C%5C%5C%5C%20slope%20%3D%20m%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B12-%28-4%29%7D%7B6-%28-2%29%7D%5Cimplies%20%5Ccfrac%7B12%2B4%7D%7B6%2B2%7D%5Cimplies%20%5Ccfrac%7B16%7D%7B8%7D%5Cimplies%202%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%28-4%29%3D2%5Bx-%28-2%29%5D%20%5C%5C%5C%5C%5C%5C%20y%2B4%3D2%28x%2B2%29%5Cimplies%20y%2B4%3D2x%2B4%5Cimplies%20%5Cblacktriangleright%20y%3D2x%20%5Cblacktriangleleft)

Answer:
B) slope of the line
;Y- intercept is none
slope of the line 
Y- intercept is none
Step-by-step explanation:
<u><em>Explanation:-</em></u>
Given equation of the line 
The equation of the straight line in slope - intercept form
y = m x + c

slope of the line 
Y- intercept is none
Answer:
hi there the diagram is 4 units
Step-by-step explanation:
65+2i is the answer- I am almost 100 percent sure
Answer:
s = 4
Step-by-step explanation:
Note that the triangle has 2 congruent base angles, both 45°
This means the triangle is isosceles with congruent legs, then
s = 4