SOLUTION: Find the equation of the line passing through the points (3, 5) and (2.8, 6).
Here is what I did:
(x1 is 3, y1 is 5), (x2 is 2.8, y2 is 6)
A.
Slope = y2 - y1/x2 - x1
Algebra.Com
Question 575955: Find the equation of the line passing through the points (3, 5) and (2.8, 6).
Here is what I did:
(x1 is 3, y1 is 5), (x2 is 2.8, y2 is 6)
A.
Slope = y2 - y1/x2 - x1
Slope = 6 - 5/2.8 - 3
= 1/-0.2
= -5
B.
y = mx + b
I picked ordered pair (3, 5) to find the y-intercept.
5 = -5(3) + b
5*-5 = -5*3 + b
-25 = -15 + b
-25 - (-15) + b
= -10
C. The equation is y = -5x - 10
Let me know if I did it correct, and if it is incorrect please do explain me how it works. Thank you.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
You have the correct slope, you just made an error in finding the y-intercept
y = mx+b
y = -5x + b
5 = -5(3) + b
5 = -15 + b
5 + 15 = b
20 = b
b = 20
So the equation is y = -5x + 20
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