What is an equation of the line that passes through the point left bracket, minus, 3, comma, 1, right bracket(−3,1) and is perpendicular to the line x, minus, 3, y, equals, 6x−3y=6?

Answer 1

To find the equation of the line that passes through the point (-3,1) and is perpendicular to the line x-3y=6, we first need to find the slope of the given line.

Given line: x-3y=6
Rearranging the equation into slope-intercept form (y = mx + b) we get:
3y = x - 6
y = (1/3)x - 2

The slope of the given line is 1/3. The slope of a line perpendicular to this line will be the negative reciprocal of 1/3, which is -3.

Now, we have the slope (-3) and a point (-3,1) that the line passes through. We can use the point-slope form of the equation of a line to find the equation of the perpendicular line:

y - y1 = m(x - x1)
y - 1 = -3(x + 3)
y - 1 = -3x - 9
y = -3x - 8

Therefore, the equation of the line that passes through the point (-3,1) and is perpendicular to the line x-3y=6 is y = -3x - 8.