To find a line perpendicular to the given equation y = (1/3)x - 2 that passes through the point (6, -9), we need to determine the slope of the given equation and then find the negative reciprocal of that slope.
The given equation is in slope-intercept form, y = mx + b, where m represents the slope. In this case, the slope is 1/3.
To find the negative reciprocal of 1/3, we invert the fraction and change its sign. Therefore, the negative reciprocal is -3.
Now that we have the slope of the perpendicular line, we can use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) represents the coordinates of the given point.
Plugging in the values from the given point (6, -9) and the slope -3, we get:
y - (-9) = -3(x - 6)
Simplifying further:
y + 9 = -3x + 18
Now, let's rearrange the equation to slope-intercept form:
y = -3x + 18 - 9
y = -3x + 9
Therefore, the equation of the line perpendicular to y = (1/3)x - 2 that includes the point (6, -9) is y = -3x + 9.