to get the equation of any straight line we simply need two points off of it, hmmm let's use the points in the picture below.
![(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-8})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{-8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-8}-\stackrel{y1}{(-8)}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{(-1)}}}\implies \cfrac{-8+8}{5+1}\implies \cfrac{0}{6}\implies 0](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B-8%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B-8%7D%29%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B-8%7D-%5Cstackrel%7By1%7D%7B%28-8%29%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B5%7D-%5Cunderset%7Bx_1%7D%7B%28-1%29%7D%7D%7D%5Cimplies%20%5Ccfrac%7B-8%2B8%7D%7B5%2B1%7D%5Cimplies%20%5Ccfrac%7B0%7D%7B6%7D%5Cimplies%200)
![\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-\stackrel{y_1}{(-8)}=\stackrel{m}{0}(x-\stackrel{x_1}{(-1)}) \implies y+8=0(x+1) \\\\\\ y+8=0x+0\implies y=\stackrel{m}{0} x\underset{b}{-8}\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}](https://tex.z-dn.net/?f=%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-%5Cstackrel%7By_1%7D%7B%28-8%29%7D%3D%5Cstackrel%7Bm%7D%7B0%7D%28x-%5Cstackrel%7Bx_1%7D%7B%28-1%29%7D%29%20%5Cimplies%20y%2B8%3D0%28x%2B1%29%20%5C%5C%5C%5C%5C%5C%20y%2B8%3D0x%2B0%5Cimplies%20y%3D%5Cstackrel%7Bm%7D%7B0%7D%20x%5Cunderset%7Bb%7D%7B-8%7D%5Cimpliedby%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D)
Answer:
Step-by-step explanation:
To prove a quadrilateral a parallelogram we prove,
1). Length of opposite sides are equal.
2). Slopes of the opposite sides are same.
Length of AB = ![\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
= ![\sqrt{(2+1)^2+(-5+2)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%282%2B1%29%5E2%2B%28-5%2B2%29%5E2%7D)
= ![3\sqrt{2}](https://tex.z-dn.net/?f=3%5Csqrt%7B2%7D)
Length of BC = ![\sqrt{(2-1)^2+(-5+2)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%282-1%29%5E2%2B%28-5%2B2%29%5E2%7D)
= ![\sqrt{10}](https://tex.z-dn.net/?f=%5Csqrt%7B10%7D)
Length of CD = ![\sqrt{(1+2)^2+(-2-1)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%281%2B2%29%5E2%2B%28-2-1%29%5E2%7D)
= ![3\sqrt{2}](https://tex.z-dn.net/?f=3%5Csqrt%7B2%7D)
Length of AD = ![\sqrt{(-1+2)^2+(-2-1)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-1%2B2%29%5E2%2B%28-2-1%29%5E2%7D)
= ![\sqrt{10}](https://tex.z-dn.net/?f=%5Csqrt%7B10%7D)
Therefore, AB = CD and BC = AD (Opposite sides are equal in length)
Slope of AB = ![\frac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
= ![\frac{-5+2}{2+1}](https://tex.z-dn.net/?f=%5Cfrac%7B-5%2B2%7D%7B2%2B1%7D)
= -1
Slope of BC = ![\frac{-5+2}{2-1}](https://tex.z-dn.net/?f=%5Cfrac%7B-5%2B2%7D%7B2-1%7D)
= -3
Slope of CD = ![\frac{1+2}{-2-1}](https://tex.z-dn.net/?f=%5Cfrac%7B1%2B2%7D%7B-2-1%7D)
= -1
Slope of AD = ![\frac{-2-1}{-1+2}](https://tex.z-dn.net/?f=%5Cfrac%7B-2-1%7D%7B-1%2B2%7D)
= -3
Slope of AB = slope of CD and slope of BC = slope AD
Therefore, AB║CD and BC║AD
Hence ABCD is a parallelogram
Answer:
no
Step-by-step explanation:
k
=
0.69942196