Select the equation of the line that passes through the point (2, –1) and has slope 3 in point-slope form.
A.
(y – 1) = 3(x + 2)
B.
(y + 1) = 3(x – 2)
C.
(x + 2) = 3(y – 1)
D.
(x – 2) = 3(y + 1)
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jimthompson5910 (jim_thompson5910):
hint: point slope form in general is
\[\Large y - y_{1} = m(x - x_{1})\]
jimthompson5910 (jim_thompson5910):
m is the slope
\(\Large (x_{1}, y_{1})\) is the point the line passes through
OpenStudy (keeli_grace):
Thank you @jim_thompson5910
But I don't understand any of that. I am a straight A student except for Algebra.
Do you think that you could explain farther, please?
jimthompson5910 (jim_thompson5910):
what's the given slope of this line?
OpenStudy (keeli_grace):
3?
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jimthompson5910 (jim_thompson5910):
So m = 3
jimthompson5910 (jim_thompson5910):
We then plug m = 3 into
\[\Large y - y_{1} = m(x - x_{1})\]
to get
\[\Large y - y_{1} = 3(x - x_{1})\]
jimthompson5910 (jim_thompson5910):
What is the given point the line passes through?
OpenStudy (keeli_grace):
2, -1?
jimthompson5910 (jim_thompson5910):
(2,-1)
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jimthompson5910 (jim_thompson5910):
So (x1,y1) = (2,-1)
what is x1 equal to?
OpenStudy (keeli_grace):
2?
jimthompson5910 (jim_thompson5910):
x1 = 2
jimthompson5910 (jim_thompson5910):
plug this into
\[\Large y - y_{1} = 3(x - x_{1})\]
to get
\[\Large y - y_{1} = 3(x - 2)\]
jimthompson5910 (jim_thompson5910):
y1 = -1
plug that into
\[\Large y - y_{1} = 3(x - 2)\]
to get
\[\Large y - y_{1} = 3(x - 2)\]
\[\Large y - (-1) = 3(x - 2)\]
\[\Large y + 1 = 3(x - 2)\]
Notice how the sign changed
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