Rule of correspondance for relations I do know how to do it (mostly) but I am having trouble with a few. I guess the question is can I use negative numbers to solve them? {(x,y): (1,2),(2,5), (3,8), , ,}
Can you tell if the relation is linear?
You mean going in a straight line on a graph?
Yes. Can you tell if its graph is a straight line?
my graph looks awful but yes it seems to be straight or linear.
Ok. I'll tell you how you can tell if it is a line without graphing. A linear relation has the following. For an equal difference in x, there must equal difference in y.
I'll explain how that is used.
You have only 3 points given. (1, 2), (2, 5), (3, 8)
Look at the first two points: (1, 2) and (2, 5) The difference in x-coordinates is 2 - 1 = 1 The difference in y-coordinates is 5 - 2 = 3 Ok so far?
Well I also noticed the pattern it had. one right, three up. if that's what you mean.
Exactly. That shows the same slope between points which means a line. Great observation.
:D
The difference in x was 1 and the difference in y was 3. Now let's look at the last two points. (2, 5), (3, 8) The difference is x is 3 - 2 = 1 The difference in y is 8 - 5 = 3 Again we have a difference in x of 1 and a difference in y of 3. Since for the same difference in x (1), we get teh same difference in y(3), we are dealing with a straight line.
Now that we know we have a straight line, we can find the equation of the line.
Do you know how to find the equation of a line given two points?
No
Not really
Ok, no problem. I'll explain it to you. There are several methods, but here is one of them.
The two-point equation of a line A line that passes through points \((x_1, y_1)\) and \((x_2, y_2)\) has equation \(y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)\)
You can use any two points of your three given points. Let's use the first two points (1, 2) and (2, 5)
sorry my tracad is messing around. but the answer appears to be 3 over 1?
\(y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)\) \(y - 2 = \dfrac{5 - 2}{2 - 1}(x - 1)\) \(y - 2 = \dfrac{3}{1}(x - 1)\) \(y - 2 = 3(x - 1)\) \(y - 2 = 3x - 3)\) \(y = 3x - 1\) The equation gives you the rule of the relation. Take the x value, multiply by 3, and subtract 1. Try it with your 3 points, and you'll see that it works.
Ok I see what you have there. If I try one with a different equation can you check to see I'm doing it right? Also can this only be done with linear equations?
@mathstudent55 (2,3) (4,4) (6,5) y=x-1? or did I miss something?
Actually that can't be right...
\(y = \dfrac{1}{2}x + 2\)
Sorry, but gtg.
ok thanks though!
yw
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