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Old 01-11-2005, 06:27 PM   #1 (permalink)
 
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Default Slope Intercept Equations

Guys...I think I went over this some time ago in math, but it has come back up to haunt me. When I did them it was easy but then I forgot and now I am back to doing them again (because for some lame reason they are on my final).

Ok, if anyone can give me the steps and the answer to this equation that was be awesome, this is on a study guide and there are a couple others like this so I need to know how to do them.

"Write the slope intercept form of the equation of the line passing through point (5 , 2) and paralell to the to the line: y = -5x - 8


If anyone can help THX VERY MUCH!
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Old 01-11-2005, 07:05 PM   #2 (permalink)
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I haven't taken algebra for several years, but here goes:
Use the Point slope formula:

1) y = m(x-x1) + y1
2) y = -5(x-5) + 2
3) Distribute
4) y = -5x+25+2
5)

Therefore the equation is y = -5x+27, or I prefer it as 27-5x.
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Old 01-11-2005, 07:06 PM   #3 (permalink)
 
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you need to use "slope-intercept" form which is y=mx+b
i cant solve it for you, but thats all i know

good luck
pccc
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Old 01-11-2005, 07:10 PM   #4 (permalink)
 
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Ok, to solve this problem, it's easiest to start off by putting it into the point-slope form. The equation of a line in p-s form is (y-y1) = m(x-x1) where (x1, y1) is a point on the line and m is the slope.

So you're given a point on the line in the question. From that you get the equation to (y-2)=m(x-5).

Since the line y=-5x-8 is parallel to it, that means the line you're finding the euqation of has the same slope as this line. Therefore the slope of the line you're finding is -5.

From that you get the equation of the line in p-s form as (y-2)=-5(x-5). From there you slove for Y and you've got it solved.

(The answer to this one is y=-5x+27)
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Old 01-11-2005, 07:16 PM   #5 (permalink)
 
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Ok, it has been a while for me to, but I'll give it a shot.
First, you get the line for y=-5x-8.
Second, you graph the coordinates that you were given(5, 2).
Third, you use the same slope from the original equation, so you would, from 5, 2, go back 5 then down one. You draw the line and it should be parallel.

My answer was: y=-5x+1, but again, I haven't done this in a while so I might be right or wrong...anyone else gat that?
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Old 01-11-2005, 07:25 PM   #6 (permalink)
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-5x+27 is the correct answer, even after checking on my calculator (see attachment).

On the left is -5x-8 and on the right is -5x+27. The resulting equation is parallel and passes through the point (5,2).
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Old 01-11-2005, 07:28 PM   #7 (permalink)
 
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Quote:
Originally posted by bla!!
Ok, to solve this problem, it's easiest to start off by putting it into the point-slope form. The equation of a line in p-s form is (y-y1) = m(x-x1) where (x1, y1) is a point on the line and m is the slope.

So you're given a point on the line in the question. From that you get the equation to (y-2)=m(x-5).

Since the line y=-5x-8 is parallel to it, that means the line you're finding the euqation of has the same slope as this line. Therefore the slope of the line you're finding is -5.

From that you get the equation of the line in p-s form as (y-2)=-5(x-5). From there you slove for Y and you've got it solved.

(The answer to this one is y=-5x+27)
I have changed it into the point slope on one of them, but the answer still came out wrong for whatever reason.


bla!!! can you look at this equation and tell me if it is right plz?

"Write the slope intercept form of the equation of the line passing through the point ( -6, 4) and paralell to the line: y = 1/2x - 3 "

My answer was " y = 1/2x + 7 "

Is that right?

And also, if it is perpendicular would this mean that it is the opposite inverse? Or what...
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Old 01-11-2005, 07:31 PM   #8 (permalink)
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Perpendicular = change the slope (m) to the opposite inverse (negative reciprocal), and solve for the y intercept the normal way.

1/2x + 7 is the correct answer.
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Old 01-11-2005, 07:33 PM   #9 (permalink)
 
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Yay, thx guys.

It's been a while since algebra 1 and my head is filled with things right now.

THX for the HELP x10!!!!!!!!!!!!
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Old 01-11-2005, 07:47 PM   #10 (permalink)
 
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Yup, that's perfect.

For the perpendicular it's the same, just for the slope you just take the inverse of the perpedicular like (1/m) and multiply by -1
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