Answer:
See explanation
Step-by-step explanation:
a) To prove that DEFG is a rhombus, it is sufficient to prove that:
- All the sides of the rhombus are congruent:
![|DG|\cong |GF| \cong |EF| \cong |DE|](https://tex.z-dn.net/?f=%7CDG%7C%5Ccong%20%7CGF%7C%20%5Ccong%20%7CEF%7C%20%5Ccong%20%7CDE%7C)
- The diagonals are perpendicular
Using the distance formula; ![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
![|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}](https://tex.z-dn.net/?f=%7CDG%7C%3D%5Csqrt%7B%280-%28-a-b%29%29%5E2%2B%280-c%29%5E2%7D)
![\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}](https://tex.z-dn.net/?f=%5Cimplies%20%7CDG%7C%3D%5Csqrt%7Ba%5E2%2Bb%5E2%2Bc%5E2%2B2ab%7D)
![|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}](https://tex.z-dn.net/?f=%7CGF%7C%3D%5Csqrt%7B%28%28a%2Bb%29-0%29%5E2%2B%28c-0%29%5E2%7D)
![\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}](https://tex.z-dn.net/?f=%5Cimplies%20%7CGF%7C%3D%5Csqrt%7Ba%5E2%2Bb%5E2%2Bc%5E2%2B2ab%7D)
![|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}](https://tex.z-dn.net/?f=%7CEF%7C%3D%5Csqrt%7B%28%28a%2Bb%29-0%29%5E2%2B%28c-2c%29%5E2%7D)
![\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}](https://tex.z-dn.net/?f=%5Cimplies%20%7CEF%7C%3D%5Csqrt%7Ba%5E2%2Bb%5E2%2Bc%5E2%2B2ab%7D)
![|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}](https://tex.z-dn.net/?f=%7CDE%7C%3D%5Csqrt%7B%280-%28-a-b%29%29%5E2%2B%282c-c%29%5E2%7D)
![\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}](https://tex.z-dn.net/?f=%5Cimplies%20%7CDE%7C%3D%5Csqrt%7Ba%5E2%2Bb%5E2%2Bc%5E2%2B2ab%7D)
Using the slope formula; ![m=\frac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
The slope of EG is ![m_{EG}=\frac{2c-0}{0-0}](https://tex.z-dn.net/?f=m_%7BEG%7D%3D%5Cfrac%7B2c-0%7D%7B0-0%7D)
![\implies m_{EG}=\frac{2c}{0}](https://tex.z-dn.net/?f=%5Cimplies%20m_%7BEG%7D%3D%5Cfrac%7B2c%7D%7B0%7D)
The slope of EG is undefined hence it is a vertical line.
The slope of DF is ![m_{DF}=\frac{c-c}{a+b-(-a-b)}](https://tex.z-dn.net/?f=m_%7BDF%7D%3D%5Cfrac%7Bc-c%7D%7Ba%2Bb-%28-a-b%29%7D)
![\implies m_{DF}=\frac{0}{2a+2b)}=0](https://tex.z-dn.net/?f=%5Cimplies%20m_%7BDF%7D%3D%5Cfrac%7B0%7D%7B2a%2B2b%29%7D%3D0)
The slope of DF is zero, hence it is a horizontal line.
A horizontal line meets a vertical line at 90 degrees.
Conclusion:
Since
and
, DEFG is a rhombus
b) Using the slope formula:
The slope of DE is ![m_{DE}=\frac{2c-c}{0-(-a-b)}](https://tex.z-dn.net/?f=m_%7BDE%7D%3D%5Cfrac%7B2c-c%7D%7B0-%28-a-b%29%7D)
![m_{DE}=\frac{c}{a+b)}](https://tex.z-dn.net/?f=m_%7BDE%7D%3D%5Cfrac%7Bc%7D%7Ba%2Bb%29%7D)
The slope of FG is ![m_{FG}=\frac{c-0}{a+b-0}](https://tex.z-dn.net/?f=m_%7BFG%7D%3D%5Cfrac%7Bc-0%7D%7Ba%2Bb-0%7D)
![\implies m_{FG}=\frac{c}{a+b}](https://tex.z-dn.net/?f=%5Cimplies%20m_%7BFG%7D%3D%5Cfrac%7Bc%7D%7Ba%2Bb%7D)
If we identify the lines by ...
1. y = -0.5x -3
2. y = -3.5x -15
3. y = 5x +19
4. y = 1.25x +4
Then the matches are ...
BC - 2
DE - 1
FG - none
HI - 4
JK - none
LM - 3
_____
Whenever there is a lot of repetitive math to do, I usually find it easier to let a calculator or spreadsheet do it. Here, the graphing calculator is computing the slope between the points. The slopes of parallel lines are the same.
Answer:
it does not matter
Step-by-step explanation:
Because you still have to put both numbers but it still does not matter what goes first as long as it is on the graph it is alright to use whichever steps feel first but x is more commonly used before y