Sure, I can help you with that. Here are the equations of the lines you requested:
1. Since the line is parallel to y = 3x + 4, it has the same slope of 3. Using a point-slope form, the equation of the line passing through P1(-2, 5) is:
y - 5 = 3(x + 2)
y - 5 = 3x + 6
y = 3x + 11
Therefore, the equation of the line is y = 3x + 11.
2. Since the line is parallel to the line passing through P1(0, -3) and P2(-1, 4), it has the same slope as that line. The slope of the line passing through P1(0, -3) and P2(-1, 4) is:
slope = (4 - (-3)) / (-1 - 0) = 7 / (-1) = -7
Using a point-slope form, the equation of the line passing through P3(-1, 4) is:
y - 4 = -7(x + 1)
y - 4 = -7x - 7
y = -7x - 3
Therefore, the equation of the line is y = -7x - 3.
3. Since the line is perpendicular to the line y = x + 4, its slope is the negative reciprocal of the slope of y = x + 4, which is -1. Using a point-slope form, the equation of the line passing through (-2, -6) is:
y - (-6) = -1(x - (-2))
y + 6 = -x - 2
y = -x - 8
Therefore, the equation of the line is y = -x - 8.
4. Since the line is perpendicular to the line passing through points P1(3, 2) and P2(4, -6), its slope is the negative reciprocal of the slope of that line. The slope of the line passing through P1(3, 2) and P2(4, -6) is:
slope = (-6 - 2) / (4 - 3) = -8
The negative reciprocal of -8 is 1/8, so the slope of the line we're looking for is 1/8. Using a point-slope form, the equation of the line passing through P3(-6, 4) is:
y - 4 = (1/8)(x - (-6))
y - 4 = (1/8)x + 3/4
y = (1/8)x + 19/4
Therefore, the equation of the line is y = (1/8)x + 19/4.
