We want to find a line such that it passes through a given point and is perpendicular to another line.
We will find that the equation of the line is: y = (1/6)*x - 3
First, we do know that for a line of the form:
y = a*x + b
where a is the slope and b is the y-intercept.
Any perpendicular line will have a slope equal to the opposite of the inverse, so the slope of the perpendicular line will be: -(1/a).
Now we want to find a perpendicular line to:
6*x + y = 2
We rewrite it in the slope-intercept form:
y = -6*x + 2
The slope is -6, then the slope of a perpendicular line will be:
-(1/-6) = (1/6)
Then the general perpendicular line is:
y = (1/6)*x + b
To find the value of b, we use the fact that our line must pass through the point (6, -2). This means that when x = 6, y = -2
Repacing that in the equation we get:
-2 = (1/6)*6 + b
-2 = 1 + b
-2 - 1 = -3 = b
Then the equation of the line is:
y = (1/6)*x - 3
If you want to learn more, you can read:
https://brainly.com/question/19954935
