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Mathematics 15 Online
OpenStudy (anonymous):

Find an equation of the line containing the point (–2, –5) and parallel to the line 4x–5y=13

OpenStudy (tkhunny):

Have you considered calculating the slope of the given line?

OpenStudy (anonymous):

honestly i really dont even know where to start

OpenStudy (tkhunny):

I do not believe that. Have you heard of the "Slope Intercept Form"?

OpenStudy (anonymous):

no i havent :/

OpenStudy (tkhunny):

How about "Point-Slope" form?

OpenStudy (anonymous):

yes ive heard of that one

OpenStudy (tkhunny):

Can you write it, here? What does it look like and what does it mean?

OpenStudy (anonymous):

y – y1 = m(x – x1) When i have a point from the graph and plug it into the slope

OpenStudy (anonymous):

am i right?

OpenStudy (tkhunny):

Yes. Very good. Can you put the GIVEN line into this form?

OpenStudy (anonymous):

ok so is it?:y– –5= 5/4 (x– –2)

OpenStudy (tkhunny):

1) We don't know that the given line passes through (-2,-5). Most likely, it doesn't. 2) We actually don't care, since all we needed was the slope, 4/5 (be more careful). 3) Using the Point-Slope Form, we have... (y-(-5)) = (4/5)(x-(-2)) or (y+5) = (4/5)(x+2)

OpenStudy (anonymous):

oh ok thanks... so what would be the next step?

OpenStudy (tkhunny):

None. It is done.

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