The equation of the line passing through point P(6, 1) and perpendicular to a line with a generic slope m is found by taking the negative reciprocal of m as the new slope. If m is 3, the slope of the perpendicular line will be -1/3. Thus, the equation is y = -1/3x + 3.
To determine the equation of a line passing through the point P(6, 1) that is perpendicular to line MN, we first need the slope of MN. Since there is no information provided about line MN in this context, let's assume the line has a generic slope, which we'll call m. For our new line to be perpendicular to MN, its slope (m') must be the negative reciprocal of m. Let's say the slope of MN is 3, as suggested by Figure A1 in your reference material. Therefore, the slope of the perpendicular line would be -1/3. An equation of a line can be expressed in the form y = mx + b, where m is the slope and b is the y-intercept.
Using point P(6, 1) and slope m' of -1/3, we substitute in the point-slope formula:
y - y1 = m'(x - x1)
y - 1 = -1/3(x - 6)
y - 1 = -1/3x + 2
y = -1/3x + 3
The final equation of the line that passes through point P and is perpendicular to line MN with a slope of 3 is y = -1/3x + 3.