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 August 9th, 2013, 02:09 PM #1 Newbie   Joined: Aug 2013 Posts: 24 Thanks: 0 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!
 August 9th, 2013, 03:36 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: pre-calculus gradient of a line a) The gradient or slope of a line is given by: $m=\frac{\Delta y}{\Delta x}$ In this problem, we are given: $m=\frac{5}{2},\,\Delta y=b-2,\,\Delta x=6-a$ and so, we may write: $\frac{b-2}{6-a}=\frac{5}{2}$ To find a pair of numbers $(a,b)$, we would solve for $b$ to obtain: $b=\frac{5}{2}(6-a)+2$ Now, choose any number $a$ where $a\ne6$, then compute $b$ 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
 August 10th, 2013, 01:24 AM #3 Newbie   Joined: Aug 2013 Posts: 24 Thanks: 0 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?
 August 10th, 2013, 02:07 AM #4 Newbie   Joined: Aug 2013 Posts: 24 Thanks: 0 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?
 August 10th, 2013, 08:00 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: pre-calculus gradient of a line Yes, that's one possible set of points. If you want integral coordinates (lattice points), let $a$ be an even integer.
 August 10th, 2013, 03:53 PM #6 Newbie   Joined: Aug 2013 Posts: 24 Thanks: 0 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?
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?

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.

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.

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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