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Question 852590: I am working on linear equations and am getting so confused when they used fractions! I have encountered an ordered pairs problem: the ordered pairs are (6,34), (13,82) using formula m= p2-p1 over h2-h1: here is some of the problem: p-34=48/7 (h-6), then p-34= 48/7h-288/7, then p=48/7h-50/7.....I just do not understand or see how they got these problems and fractions...I hope you can explain this to me, or show me! Thank you!
Found 2 solutions by josgarithmetic, LinnW: Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! The way you are handling the numbers and variables for some of them conforms to point-slope form for a line. Your general points are in the form, (p,h).
m is often used as a variable for slope. For lines in the plane, m, slope, is vertical change divided by horizontal change. That is, for the variables you are using, change in p divided by change in h.
Look at the value for m for the two points you were given:




--------Value of the slope of those two points.
Look again at the first equation giving the expression for the slope.

You KNOW where the values used in the formula came from. You see a specific difference between p values and this is divided by a specific difference between h values.
CLEAR the denominator by multiplying left side and right side by (6-13). You will obtain:

-------both symmetric property and commutative prop. of multipl.
The "13" and the "82" are a specific point, and you already know . The "6" and the "34" come from (6, 34), but this could just as well be any general point on the line defined by the two given points. Meaning, that more broadly than (6, 34), the point is any (h, p).
Then the equation can be stated as:

You know what m is...

But you can use ANY point on the line defined by (6,34) and (13,82); Since both these points are on this line, you can also say:
The equation form is POINT-SLOPE FORM. It allows to express a line using the slope of the line and a point on the line. .... ANY point on the line.
Answer by LinnW(1048) (Show Source):
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