Posted by **Katty** on Tuesday, July 24, 2007 at 12:39pm.

how do I write the equation of a line perpendicular to the line 6x – 3y=10 and containing the point (-6, 2).

The line you are given can have its equation rewritten im y = mx + b as:

y = 2x - 10/3

That tells you that is slope is m = 2. A line that is perpendicular to that must have a slope of -1/2, because the product of slopes of perpendicular lines must be m = -1. (Trust me on that one, or check your textbook.)

A new perpendicular line going through a point (x*,y*) will have the equation

y - y* = m (x - x*)

y - 2 = (-1/2)(x -(-6))

y - 2 = -x/2 - 3

y = -(x/2) -1

(y - )/(

Ignore the last line of my last answer. It's some leftover garbage I neglected to delete