Line (2)

Problem:



Solution:

The line $ 3x + 7y = 1 $ in slope intercept form is given by

$ y = \frac{-3}{7}x + \frac{1}{7} $.

Therefore, the slope of the line is $ \frac{-3}{7} $.

The line perpendicular to it must have slope $ \frac{7}{3} $.

Therefore, the required line must be, by point-slope form, given by

$ y - 8 = \frac{7}{3}(x - 1) $

Simplifying to give the point slope form as follow

$ y - 8 = \frac{7}{3}x - \frac{7}{3} $
$ y = \frac{7}{3}x + \frac{17}{3} $

The standard form can be found by putting $ y $ to the right hand side, that gives.

$ 0 = \frac{7}{3}x - y + \frac{17}{3} $.

The solution can be made prettier by multiplying 3 throughout, that gives

$ 7x - 3y + 17 = 0 $