The formula for finding the slope of a line is $y_2-y_1\over x_2-x_1$, but can it be reversed into $y_1-y_2\over x_1-x_2$ and still be right?
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6$\begingroup$ Yep, just multiply by $1 = \frac{-1}{-1}$ $\endgroup$– Krishan BhallaMay 2, 2015 at 23:10
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2$\begingroup$ To expand on KB94's comment, this works because $-(a - b) = -a - (-b) = -a + b = b - a$; this is happening in both the top and bottom, when multiplying by $-1/-1$. $\endgroup$– pjs36May 2, 2015 at 23:42
3 Answers
More generally, the line through $(x_1, y_1)$ and $(x_2, y_2)$ has slope $\frac{y_2-y_1}{x_2-x_1} $.
Swapping the points, the line through $(x_2, y_2)$ and $(x_1, y_1)$ has slope $\frac{y_1-y_2}{x_1-x_2} $.
But $\frac{y_2-y_1}{x_2-x_1} =\frac{y_1-y_2}{x_1-x_2} $, so the two lines have the same slope.
Yes, I think it's right. Here's an example for (8, 5) and (4, 2):$$2-5\over 4-8$$$$-3\over-4$$$$3\over4$$and now it's time to reverse it:$$5-2\over8-4$$$$3\over4$$I can now conclude that I can reverse it for it to be right (except when the numerator or denominator is zero because division by zero is undefined).
sure , for example the points (4,8) and (8,2) if (4,8) = (x1,y1) and (8,2)=(x2,y2) m=y2-y1/x2-x1 = -6/4 = -3/2
also m=y1-y2/x1-x2 = 6/-4 = -3/2 .
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$\begingroup$ Please use MathJax to format math, so they can appear cleaner. $\endgroup$ May 13, 2015 at 3:07