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 March 27th, 2014, 04:41 AM #1 Newbie   Joined: Mar 2014 Posts: 6 Thanks: 0 find the equations Hello, hopefully I'm on the right bit I need to find the equations of a few plots first is: A(-1,-4) B(5-2) I'd like to know if this is the correct way: m= -2-4/5-1 = -6/4 =-3/2 so m= -3/2 -4=-3/2x + c c=-11/2 so the equation is y=-3/2x-11/2
 March 27th, 2014, 04:56 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 Take care to include all the necessary symbols. For example, did you mean "B(5-2)" to be "B(5,2)" or "B(5,-2)"?
 March 27th, 2014, 05:00 AM #3 Newbie   Joined: Mar 2014 Posts: 6 Thanks: 0 Re: find the equations ah sorry mean B(5,-2)
March 27th, 2014, 05:27 AM   #4
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Re: find the equations

Quote:
 Originally Posted by mintedm Hello, hopefully I'm on the right bit I need to find the equations of a few plots first is: A(-1,-4) B(5-2) I'd like to know if this is the correct way: m= -2-4/5-1 = -6/4 =-3/2 so m= -3/2
No, it's not. The slope of the line through $(x_0, y_0)$ and $(x_1, y_1)$ is given by
$\frac{y_1- y_0}{x_1- x_0}$ or $\frac{y_0- y_1}{x_0- x_1}$

That is what you are trying to do but you are being careless with signs!
$m= \frac{-2- (-4)}{5- (-1)}= \frac{-2+ 4}{5+ 1}= \frac{2}{6}= \frac{1}{3}$

$-4=-3/2x + c$
I presume the "-4" is from (-1, -4) so x should be -1: -4= (1/3)(-1)+ c so c= -4+ 1/3= -12/3+ 1/3= -11/3

Quote:
 c=-11/2 so the equation is y=-3/2x-11/2
No, it should be y= (1/3)x- 11/3. That, of course, is the same as 3y= x- 11 or x- 3y= 11.

Another way of doing this whole problem is to recognize that any (non-vertical) line can be written as y= mx+ c for some numbers m and c. Since the point (-1, -4) is on the line, -4= m(-1)+ c. Since (5, -2) is on the line, -2= m(5)+ c.

Eliminate c from the equations by subtracting the first equation from the second: (-2- (-4))= 5m+ c- (-m+ c) or (-2+ 4)= 2= 6m so m= 2/6= 1/3. You should see that the calculations there are exactly the same as $m= \frac{-2- (-4)}{5- (-1)}= \frac{-2+ 4}{5+ 1}= \frac{2}{6}= \frac{1}{3}$ before. Since m= 1/3, -2= (1/3)(5)+ c so c= -2- 5/3= -6/2- 5/3= -11/3 OR -4= (1/3)(-1)+ c so c= -4+ 1/3= -12/3+ 1/3= -11/3. In either case, y= (1/3)x- 11/3 again.

 March 27th, 2014, 05:36 AM #5 Newbie   Joined: Mar 2014 Posts: 6 Thanks: 0 Re: find the equations Explains a lot, thanks. Going to have another shot at it with your help, and see if I can get it correct this time before moving on.
 March 27th, 2014, 07:27 AM #6 Newbie   Joined: Mar 2014 Posts: 6 Thanks: 0 Re: find the equations ok next one P(1,3) Q(4,2) $\dfrac{2-3}{4-1}$ = $\dfrac{-1}{3}$= $-\dfrac{1}{3}$ m=$-\dfrac{1}{3}$ 3= ($-\dfrac{1}{3}$x)+c 3+($-\dfrac{1}{3}$) = $\dfrac{2}{3}$ so the equation should be y=$\dfrac{1}{3}$x+$\dfrac{2}{3}$
 March 27th, 2014, 08:20 AM #7 Newbie   Joined: Mar 2014 Posts: 6 Thanks: 0 Re: find the equations edit i mean $\dfrac{8}{3}$ not 2/3
March 27th, 2014, 11:21 AM   #8
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Re: find the equations

Quote:
 Originally Posted by mintedm ok next one P(1,3) Q(4,2) $\dfrac{2-3}{4-1}$ = $\dfrac{-1}{3}$= $-\dfrac{1}{3}$ m=$-\dfrac{1}{3}$ 3= ($-\dfrac{1}{3}$x)+c 3+($-\dfrac{1}{3}$) = $\dfrac{2}{3}$ so the equation should be y=$\dfrac{1}{3}$x+$\dfrac{2}{3}$
That's quite messy, Minter...if I was your teacher, I'd deduct points even if correct

y = mx + b [1]

m = (2 - 3) / (4 - 1) = -1/3
Substitute in [1]:
y = (-1/3)x + b [2]

Using P(1,3) and substituting x=1 and y=3 in [2]:
3 = (-1/3)(1) + b
3 = -1/3 + b
b = 3 + 1/3
b = 9/3 + 1/3
b = 10/3
Substitute in [2]:
y = (-1/3)x + 10/3

You don't need to do it EXACTLY that way...but why take a chance...get my drift?

 March 27th, 2014, 11:33 AM #9 Newbie   Joined: Mar 2014 Posts: 6 Thanks: 0 Re: find the equations ok thanks Denis practise practise!!!!

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