anonymous
  • anonymous
Find the equation, in standard form, of the line passing through the points (2,-3) and (4,2).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
1) find the slope
anonymous
  • anonymous
would slope be 5/2?

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More answers

anonymous
  • anonymous
\[\frac{2-(-3)}{4-2}\] is a start
anonymous
  • anonymous
yes, you are right, it is \(\frac{5}{2}\)
anonymous
  • anonymous
so thats how you get a slope! thanks. would the answer be A then? y=5/2x+8
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
well if they weren't so mean you would be done but both A and D have slope \(\frac{5}{2}\) so don't jump the gun yet
anonymous
  • anonymous
we are going to have to pick between them
anonymous
  • anonymous
so even though D says 5x-2y=16 it is a slope like 5/2?
anonymous
  • anonymous
next comes "point - slope" formula \[y-y_1=m(x-x_1)\] you can pick whichever point you like , i pick \((4,2)\) but lets wait a sec so i can answer your last question first
anonymous
  • anonymous
\[5x-2y=16 \] if we solve for \(y\) we get \[-2y=-5x+16\] divide by \(-2\) and get \[y=\frac{5}{2}x-8\]
anonymous
  • anonymous
so yes, it also has slope \(\frac{5}{2}\) you cannot read off the slope unless it looks like \(y=mx+b\)
anonymous
  • anonymous
is that more or less clear?
anonymous
  • anonymous
oh okay!
anonymous
  • anonymous
i will take that as a "yes"
anonymous
  • anonymous
yeah that was very clear, thanks :)
anonymous
  • anonymous
between the two though, I cant figure out the answer
anonymous
  • anonymous
The answer is d
anonymous
  • anonymous
k we have \[y-y_1=m(x-x_1)\] putting in the numbers we get \[y-2=\frac{5}{2}(x-4)\]
anonymous
  • anonymous
now the steps are always the same multiply out on the right and get \[y-2=\frac{5}{2}x-10\] add 2 and get \[y=\frac{5}{2}x-8\]
anonymous
  • anonymous
so it is not A, they put that there to see if you would jump to the wrong conclusion, which you almost did it is D
anonymous
  • anonymous
thank you! :) @ask1709 @satellite73

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