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Mathematics 18 Online
OpenStudy (keeli_grace):

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)

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?

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)

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

OpenStudy (keeli_grace):

OH!!!!!!! Thank you so much!!! :)

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