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August 9th, 2013, 02:09 PM   #1
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pre-calculus gradient of a line

I'm taking a university paper on pre-calculus. I'm having difficulty with this subject as I have very little background in maths.

One of the questions I am struggling on is the following -

a) i) (a,2) and (6,b) are co-ordinates of 2 different points that lie on a straight line that has a gradient of 5/2. Note that a cannot equal 6 and b cannot equal 2. Find a value for a and b and hence a possible equation for the line.
ii) Give an explanation as to why the equation of the line you have found in (i) is only one line of many that has a gradient of 5/2 and has two points of the form (a.2) and (6,b) lying on it.
b) i) graphically or otherwise, show and explain the orientation of two lines, one line with a gradient m1 = -2/5 and the other with a gradient m2 = 5/2.
ii) Given a line with a gradient m1 = 3, find the gradient (m2) of another line that would posses the same orientation you found in (b) (i)

Any help in solving this would be much appreciated!
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August 9th, 2013, 03:36 PM   #2
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Re: pre-calculus gradient of a line

a) The gradient or slope of a line is given by:



In this problem, we are given:



and so, we may write:



To find a pair of numbers , we would solve for to obtain:



Now, choose any number where , then compute using the above formula.

I will leave you to explain why there are an infinite number of solutions.

For part b), consider the following:

http://mathhelpboards.com/math-notes...opes-2953.html
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August 10th, 2013, 01:24 AM   #3
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Re: pre-calculus gradient of a line

So when I solve for b, where a= 5,
I get -
b= 5/2 (6-a) + 2
b= 5/2 (6-5) + 2
b= 5/2 * 1 + 2
b= 5/2 * 1/1 + 2
b= 5/2 + 2
b= 5/2 + 2/1
b= 5/2 + 4/2
b= 9/2

Am I doing something wrong?
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August 10th, 2013, 02:07 AM   #4
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Re: pre-calculus gradient of a line

So if I take it a step further and 9/2 = 4.5, then the co-ordinates would be (5, 2)(6, 4.5)
m = y-y1/x-x1 =4.5-2/6-5 = 2.5/1 which is the same thing as 5/2. So I've done it right?
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August 10th, 2013, 08:00 AM   #5
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Re: pre-calculus gradient of a line

Yes, that's one possible set of points. If you want integral coordinates (lattice points), let be an even integer.
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August 10th, 2013, 03:53 PM   #6
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Re: pre-calculus gradient of a line

Would this be correct -
The equation of the line in (i) is only one line of many with the gradient of 5/2 that passes through points of the form (a,2) and (6,b) as there can be multiple parallel lines which would pass through these potential points and posses the same gradient?
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August 10th, 2013, 05:19 PM   #7
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Re: pre-calculus gradient of a line

Quote:
Originally Posted by alyssa jesse
So when I solve for b, where a= 5,
I get -
b= 5/2 (6-a) + 2
b= 5/2 (6-5) + 2
b= 5/2 * 1 + 2
b= 5/2 * 1/1 + 2
b= 5/2 + 2
b= 5/2 + 2/1
b= 5/2 + 4/2
b= 9/2
Am I doing something wrong?
WHY are you going through all those steps?
5/2 * 1 + 2
= 5/2 + 2
= 5/2 + 4/2
= 9/2
Important to cut down steps...could be a timed test...get my drift?
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August 10th, 2013, 05:46 PM   #8
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Re: pre-calculus gradient of a line

Quote:
Originally Posted by alyssa jesse
Would this be correct -
The equation of the line in (i) is only one line of many with the gradient of 5/2 that passes through points of the form (a,2) and (6,b) as there can be multiple parallel lines which would pass through these potential points and posses the same gradient?
Yes, I believe you have fully grasped the point of this part of the exercise.
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August 10th, 2013, 06:29 PM   #9
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Re: pre-calculus gradient of a line

Quote:
Originally Posted by Denis
Quote:
Originally Posted by alyssa jesse
So when I solve for b, where a= 5,
I get -
b= 5/2 (6-a) + 2
b= 5/2 (6-5) + 2
b= 5/2 * 1 + 2
b= 5/2 * 1/1 + 2
b= 5/2 + 2
b= 5/2 + 2/1
b= 5/2 + 4/2
b= 9/2
Am I doing something wrong?
WHY are you going through all those steps?
5/2 * 1 + 2
= 5/2 + 2
= 5/2 + 4/2
= 9/2
Important to cut down steps...could be a timed test...get my drift?
Hi Denis,
The reason why I am being methodical about showing my steps exactly is so someone can easily point out if I am making a mistake along the way.
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August 10th, 2013, 06:30 PM   #10
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Re: pre-calculus gradient of a line

Quote:
Originally Posted by MarkFL
Quote:
Originally Posted by alyssa jesse
Would this be correct -
The equation of the line in (i) is only one line of many with the gradient of 5/2 that passes through points of the form (a,2) and (6,b) as there can be multiple parallel lines which would pass through these potential points and posses the same gradient?
Yes, I believe you have fully grasped the point of this part of the exercise.
Thank you Mark!!
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