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Ok I need some help with my college algebra. The first problem I have to put the final equations or lines in both slope-intercept form and standard form.
The problem is: Find the slope and the right the equation of the lines in both slope-intercept form through the points of (-2,3) and (7,-3).
Call me dumb but im lost here. Anyone got any help?
1995 Patriot Red T-Topped Z28 A4<br /><br />Mystery rebuild in progress.<br /><br />Soon to have 383 ways to beat KBreezy and Shane. :D
y= -6/9x+b then plug in a point to get y int. -3= -2/3(7)+b -4.66+b= -3 so then b = 5/3 final equation is y= -2/3x + 5/3 i could be wrong its been about 3 years so double check that
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Oh god how I wish I was still doing this math. Enjoy it, and hope your major doesn't involve higher math.
<b><i>Don\'t Ever Judge Me</i></b><br /><a href=\"http://www.cardomain.com/id/earnhardt132000\" target=\"_blank\">\'96 Firebird Y87 3.8L V6 5spd</a><br />--No mods as of yet, shoot me some ideas if ya want.<br />178,000 miles and counting
Y=the second number in your coordinates (x,y)
M=the slope of the line
X=the first number in coordinates(x,y)
B=the Y intercept or where the line crosses the Y axis (vertical line)
So you have the points (-2,3) and (7,-3) and you know that the M (the slope) and B (the Y intercept) are the same.
First find the slope. You can find it by taking the rise over run. So from (-2,3) and (7,-3) it rises -6 and runs 9. -6/9= -2/3 = M
Next, plug M and one set of coordinates into the formula Y=mx+b. So, 3= (-2/3*-2)+b Now solve for "b".
Take -2/3*-2= 4/3 3=4/3+b
Subtract 4/3 from 3 to get 5/3 5/3=b
Answer is Y=-2/3x+5/3
Hope this helps. If not I would just go talk to your teacher.
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good practice.
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