Line

Problem:



Solution:

The line x + 3y = 5 can be written in slope-intercept form as follow:

$ 3y = -x + 5 $
$ y = \frac{-1}{3}x + \frac{5}{3} $

Therefore, the slope of the line is $ \frac{-1}{3} $.

The equation of the required line, by point-slope form, would then be

$ y - 6 = \frac{-1}{3}(x - (-2)) $.

Just simplify this to slope intercept form to give the answer

$ y - 6 = \frac{-1}{3}(x + 2) $.
$ y - 6 = \frac{-1}{3}x + \frac{-2}{3} $.
$ y  = \frac{-1}{3}x + \frac{16}{3} $.

Standard form can be found by just putting the $ y $ to right hand side, that gives

$ 0 = \frac{-1}{3}x - y + \frac{16}{3} $.

It could be made prettier by multiply -3 throughout, that gives

$ x + 3y - 48 = 0 $.