Straight line graphs

poochie03 · October 30th, 2011, 01:17 PM

Hi, was wondering if anyone can help please?
The point A(-6,4) and the point B lie on line L.
Find an equation for L in the form ax+by+c=0 where a, b and c are integers?
Thanks

poochie03 · October 30th, 2011, 01:34 PM

Sorry supposed to be point B is (8,-3)
Thanks

MarkFL · October 30th, 2011, 02:47 PM

First, we may compute the slope m of the line:

$m=\frac{\Delta y}{\Delta x}=\frac{-3-4}{8-(-6)}=-\frac{7}{14}=-\frac{1}{2}$

Next, using the point-slope form for a line, we find:

$y-(-3)=-\frac{1}{2}$x-8$$

$y+3=-\frac{x}{2}+4$

$\frac{x}{2}+y-1=0$

$1\cdot x+2y+(-2)=0$

poochie03 · November 5th, 2011, 01:00 PM

Thank you

October 30th, 2011, 01:17 PM	#1
poochie03 Newbie Joined: Sep 2011 Posts: 16 Thanks: 0	Straight line graphs Hi, was wondering if anyone can help please? The point A(-6,4) and the point B lie on line L. Find an equation for L in the form ax+by+c=0 where a, b and c are integers? Thanks
$poochie03 is offline$

October 30th, 2011, 01:34 PM	#2
poochie03 Newbie Joined: Sep 2011 Posts: 16 Thanks: 0	Re: Straight line graphs Sorry supposed to be point B is (8,-3) Thanks
$poochie03 is offline$

October 30th, 2011, 02:47 PM	#3
MarkFL Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs	Re: Straight line graphs First, we may compute the slope m of the line: $m=\frac{\Delta y}{\Delta x}=\frac{-3-4}{8-(-6)}=-\frac{7}{14}=-\frac{1}{2}$ Next, using the point-slope form for a line, we find: $y-(-3)=-\frac{1}{2}\(x-8\)$ $y+3=-\frac{x}{2}+4$ $\frac{x}{2}+y-1=0$ $1\cdot x+2y+(-2)=0$
$MarkFL is offline$

November 5th, 2011, 01:00 PM	#4
poochie03 Newbie Joined: Sep 2011 Posts: 16 Thanks: 0	Re: Straight line graphs Thank you
$poochie03 is offline$

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