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Mathematics 12 Online
OpenStudy (anonymous):

Write an equation of the line that passes through the point (–4, –1) and is (a) parallel to and (b) perpendicular to the line 2x + 7y = 14.

OpenStudy (anonymous):

2x + 7y = 14 first we will put it in y = mx + b form 7y = -2x + 14 y = -2/7x + 2 we will then use this formula : y - y1 = m(x - x1) One thing you need to know is that parallel lines have the same slope, and perpendicular lines have the negative reciprocal slope. So for your parallel line, your slope(m) will be -2/7, and your perpendicular slope(m) will be 7/2. are you with me so far ?

OpenStudy (anonymous):

Yes I'm with you

OpenStudy (anonymous):

I think so anyway

OpenStudy (anonymous):

(a) parallel line -- slope(m) = -2/7... points (-4,-1)..x1 = -4, y1 = -1 y - y1 = m(x - x1) y -(-1) = -2/7(x - (-4) y + 1 = -2/7 (x + 4) y + 1 = -2/7x - 1/14 y = -2/7x - 1/14 - 1 y = -2/7x - 15/14 (b) perpendicular line --slope(m) = 7/2...points (-4,-1)..x1 = -4, y1 = -1 y - y1 = m(x - x1) y - (-1) = 7/2(x - (-4) y + 1 = 7/2(x + 4) y + 1 = 7/2x + 7/8 y = 7/2x + 7/8 - 1 y = 7/2x - 1/8 do you have any questions ?

OpenStudy (anonymous):

do you understand what I just did ?

OpenStudy (anonymous):

Not Really....

OpenStudy (anonymous):

But as far as I can tell then y = -2/7x - 15/14 is parreal and y=7/2x-1/8 is perpendicular?

OpenStudy (anonymous):

To find the slope you put the equation in y = mx + b form, where m is the slope. The points will be the same (-4,-1)..x1 = -4, y1 = -1. you use the same formula for both lines. y - y1 = m(x - x1) the slopes will be different....parallel slopes are the same and perpendicular slopes are the negative reciprocal...that just means take the slope and flip it, and change the sign. Then you just plug in your info into the equation and solve for y.

OpenStudy (anonymous):

yes.....parallel is y = -2/7x - 15/14 and perpendicular is y = 7/2x - 1/8

OpenStudy (anonymous):

Thanks! I think I finally got it :)

OpenStudy (anonymous):

good....glad I could help :)

OpenStudy (anonymous):

Thanks again

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