find the equation of a line that is perpendicular to the line y=1/6x+7 and contains the point (-5,0)

y=
(Type your answer in slope intercept form)

slope has to be -6 because it is -1/(1/6)

note m' = -1/m

so of form
y = -6 x + b
put the point in
0 = -6 (-5) + b
so
b = -30
then
y = -6 x -30

To find the equation of a line that is perpendicular to another line, you need to determine the slope of the original line and then find the negative reciprocal of that slope.

Given that the equation of the original line is y = (1/6)x + 7, we can see that its slope is 1/6.

The negative reciprocal of 1/6 is -6/1 or -6.

Now that we have the slope of the perpendicular line, we can use the point-slope form of a line to find the equation. The point-slope form is given as y - y1 = m(x - x1), where m represents the slope and (x1, y1) represents a point on the line.

Using the point (-5, 0) and the slope of -6, we can substitute these values into the point-slope form:

y - 0 = -6(x - (-5))

Simplifying:

y = -6(x + 5)

Expanding:

y = -6x - 30

Therefore, the equation of the line that is perpendicular to y = (1/6)x + 7 and contains the point (-5, 0) is y = -6x - 30.