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

To write the equation of a line perpendicular to the line 6x - 3y = 10 and containing the point (-6, 2), follow these steps:

1. First, rewrite the given line in slope-intercept form (y = mx + b) by isolating y:
6x - 3y = 10
-3y = -6x + 10
y = (2x - 10/3)

2. Determine the slope of the given line. The slope (m) is the coefficient of x in the slope-intercept form. In this case, m = 2.

3. Since a line perpendicular to another line has a negative reciprocal slope, the slope of the new line will be -1/m. Therefore, the slope of the new line will be -1/2.

4. Use the point-slope form of a line to find the equation of the new line. The point-slope form is:
y - y1 = m(x - x1)

Substituting (-6, 2) for (x1, y1) and -1/2 for m:
y - 2 = (-1/2)(x - (-6))
y - 2 = (-1/2)(x + 6)
y - 2 = (-1/2)x - 3
y = (-1/2)x - 1

Therefore, the equation of the line perpendicular to 6x - 3y = 10 and containing the point (-6, 2) is y = (-1/2)x - 1.