Answer:
If the lines BC and DE are parallel, the value of c is c=-2
Step-by-step explanation:
We are given line segment BC with end points B(2, 2) and C(9,6) and line segment DE with endpoints D(c, -7) and E(5, -3).
Using slope formula:
we can find point c
When 2 lines are parallel there slope is same.
So, Slope of line BC =Slope of Line DE
![\frac{y_2-y_1}{x_2-x_1}=\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%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
We have:
![x_1=2, y_1=2, x_2=9, y_2=6 \ for \ line \ BC \ and \\\x_1=c, y_1=-7, x_2=5, y_2=-3 \ for \ line \ DE \](https://tex.z-dn.net/?f=x_1%3D2%2C%20y_1%3D2%2C%20x_2%3D9%2C%20y_2%3D6%20%5C%20for%20%5C%20line%20%5C%20BC%20%5C%20and%20%5C%5C%5Cx_1%3Dc%2C%20y_1%3D-7%2C%20x_2%3D5%2C%20y_2%3D-3%20%5C%20for%20%5C%20line%20%5C%20DE%20%5C)
Putting values and finding c
![\frac{6-2}{9-2}=\frac{-3-(-7)}{5-c}\\ \frac{4}{7}=\frac{-3+7}{5-c} \\ \frac{4}{7}=\frac{4}{5-c} \\Cross \ multiply:\\4(5-c)=4*7\\20-4c=28\\-4c=28-20\\-4c=8\\c=\frac{8}{-4}\\c=-2](https://tex.z-dn.net/?f=%5Cfrac%7B6-2%7D%7B9-2%7D%3D%5Cfrac%7B-3-%28-7%29%7D%7B5-c%7D%5C%5C%20%5Cfrac%7B4%7D%7B7%7D%3D%5Cfrac%7B-3%2B7%7D%7B5-c%7D%20%5C%5C%20%5Cfrac%7B4%7D%7B7%7D%3D%5Cfrac%7B4%7D%7B5-c%7D%20%5C%5CCross%20%5C%20multiply%3A%5C%5C4%285-c%29%3D4%2A7%5C%5C20-4c%3D28%5C%5C-4c%3D28-20%5C%5C-4c%3D8%5C%5Cc%3D%5Cfrac%7B8%7D%7B-4%7D%5C%5Cc%3D-2)
So, If the lines BC and DE are parallel, the value of c is c=-2