Hi, there--

The Problem:
Consider the line y=7x-6 
Find the equation of the line that is parallel to this line and passes through the point (6,-5). 
Find the equation of the line that is perpendicular to this line and passes through the point (6,-5).

A Solution:
When two lines are parallel, they have the same slope, but the y-intercepts are different.

{1} parallel line
Your original equation is in slope-intercept form: (y=mx+b). The slope m is the coefficient of the 
x term; it's  7.

We know that the parallel line has a slope of 7 and that it passes through the point (6,-5). 
We'll substitute these known values into the slope-intercept form and solve for the 
y-intercept b.

y = mx + b
-5 = (7)(6) + b
-5 = 42 + b
b = -47

The equation of the parallel line is y = 7x - 47.


{2} perpendicular line
If a line has a slope m, then the line perpendicular to it has the slope -1/m.

Since the original equation has a slope of 7, the perpendicular line has a slope of -1/7. 
From here the procedure is quite similar to finding the parallel line's equation. Use the
slope-intercept form to find the y-intercept.

y = mx + b
-5 = (-1/7)(6) + b
-5 = (-6/7) + b
b = -5 + 6/7
b = -29/7

The equation for the perpendicular line is y = (-1/7)x - 29/7.

Please email me if you have questions about the solution. I'd appreciate the feedback, and I'm glad 
to help.

Ms.Figgy
math.in.the.vortex@gmail.com