Asked by Mary
Find the value of k so that the line containing the points (−5,0) and (k,−2) is parallel to the line
y=(5/9)x−2
y=(5/9)x−2
Answers
Answered by
Jai
First we find the slope of the given points:
m = (y2 - y1)/(x2 - x1)
Substituting,
m = (0 - (-2))/(-5 - k)
m = 2/(-5 - k)
Then, we get the slope of the given line.
y = (5/9)x − 2
Note that it follows the form y = mx + b, where m is the slope and b is the y-intercept. Therefore, its slope = 5/9.
Recall that when two lines are parallel with each other, their slopes are equal. Thus, we equate them:
2/(-5 - k) = 5/9
2 * 9 = 5(-5 - k)
18 = -25 - 5k
5k = -25 - 18
5k = -43
k = -43/5
Hope this helps~ :3
m = (y2 - y1)/(x2 - x1)
Substituting,
m = (0 - (-2))/(-5 - k)
m = 2/(-5 - k)
Then, we get the slope of the given line.
y = (5/9)x − 2
Note that it follows the form y = mx + b, where m is the slope and b is the y-intercept. Therefore, its slope = 5/9.
Recall that when two lines are parallel with each other, their slopes are equal. Thus, we equate them:
2/(-5 - k) = 5/9
2 * 9 = 5(-5 - k)
18 = -25 - 5k
5k = -25 - 18
5k = -43
k = -43/5
Hope this helps~ :3
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