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