Thread: Functions, Graphs, and Limits

First off, really sorry if i put this in the wrong thread, its my first time on this forum. These questions are from my university calculus book, but in the review section, and for that reason, i didn't put this in the calculus section of the forum.

Write equations of the lines through the given point (a) parallel to the given line and (b) perpendicular to the given line.

1. (-3, 2) x + y = 7
2. (-2/3, 7/8) 3x + 4y = 7
3. (-1, 0) y + 3 = 0

Thanks in advance!

Hi Auger

Equation of straight line can be written as : y = mx + c, where m is the slope or gradient and c is a constant.

Do you know the relation of the slopes for the lines which are :
1. parallel ?
2. perpendicular ?

If you've figured out the relation, you can find the slope needed, then substitute it to the equation of the line along with the coordinate given to find c. So, you'll have the answer

the answer in the textbook isn't in y=mx+b form. these are the answers they give:

1. a) x+y+1=0 b) x-y+5=0
2. a) 6x+8y-3=0 b) 96x-72y+127=0
3. a) y=0 b) x+1=0

Hi Auger

It's the same. You also can write the equation like this :
y = mx + c
y -mx -c = 0

And you'll find the same answer as in your textbook

For 1. (-3, 2) x + y = 7

y=-x+7
Therefore the slope is -1 for a parallel line b/c parallel lines have = slopes.

So you use the point slope formula...
(y-2)=-1(x+3)
y=-x-1 is your first answer

for perpendicular, the slope is the negative reciprocal, which would be 1
so

y-2=1(x+3)
y=x+5

Oh ok, i know the point slope forumla i just didnt know you had to use it for this. thank you very much for you help! i have a couple other questions that i cant figure out that i might just have to post hahaha.