How would I express in standard form: contains point (-1,4) and is parallel to x-5y=5?

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- Sep 29th 2009, 07:10 PMslayingiceLinear problem
How would I express in standard form: contains point (-1,4) and is parallel to x-5y=5?

- Sep 29th 2009, 07:44 PMpickslides
I think you are looking for the line that is parallel to $\displaystyle x-5y=5$ that contains the point $\displaystyle (-1,4)$ ?

Haivng a play with this guy gives.

$\displaystyle x-5y=5$

$\displaystyle -5y=5-x$

$\displaystyle y=\frac{5}{-5}-\frac{x}{-5}$

has the gradient $\displaystyle m= \frac{1}{5}$

So the line you need to find is $\displaystyle y = \frac{1}{5}x+c $ with $\displaystyle (-1,4)$

$\displaystyle 4 = \frac{1}{5}\times -1+c $

Now solve for c - Sep 29th 2009, 10:12 PMpacman
http://www.mathhelpforum.com/math-he...40bd7de0-1.gif that contains the point http://www.mathhelpforum.com/math-he...d4f148a5-1.gif ,we have

x - 5 = 5y, divide BS by 5,

y = (1/5)x - (5/5),

y = (1/5)x - 1, - - - -**(1)**

compare it with intercept-slope form, y = mx + b

m = 1/5, if another line parallel to it passing through (-1,4), we have

4 = (1/5)(-1) + c, solving for c

4 = -1/5 + c,

4 + 1/5 = c,

(20 + 1)/5 = c,

21/5 = c

thus the line's equation is,

y = (1/5)x + 21/5 - - - -**(2)**

see the plot below . . . . JUST double CLICK the picture for a BETTER VIEW.