Final answer:
a) The equation of the line parallel to 6x+3y=9 and passing through (-2, 3) is y = -2x - 1. b) The equation of the line parallel to 5x−7y=0 and passing through (7, -6) is y = (5/7)x - 36/7. c) The equation of the line perpendicular to 6x+3y=9 and passing through (-2, 3) is y = (1/2)x - 1. d) The equation of the line perpendicular to 5x−7y=0 and passing through (7, -6) is y = (-7/5)x + 54/5.
Step-by-step explanation:
a) To find the equation of a line parallel to 6x+3y=9, we need the slope of the given line. Rearranging the equation into slope-intercept form, we get y = -2x + 3. Since parallel lines have the same slope, the parallel line also has a slope of -2. Now we can use the point-slope form, y - y1 = m(x - x1), where (x1, y1) is the given point (-2, 3), and m is the slope. Plugging in the values, we get y - 3 = -2(x + 2). Simplifying, we get y = -2x - 1. Therefore, the equation of the line parallel to 6x+3y=9 and passing through (-2, 3) is y = -2x - 1.
b) To find the equation of a line parallel to 5x−7y=0, we need to find the slope of the given line. Rearranging the equation into slope-intercept form, we get y = (5/7)x. Since parallel lines have the same slope, the parallel line also has a slope of (5/7). Now we can use the point-slope form, y - y1 = m(x - x1), where (x1, y1) is the given point (7, -6), and m is the slope. Plugging in the values, we get y - (-6) = (5/7)(x - 7). Simplifying, we get y = (5/7)x - 36/7. Therefore, the equation of the line parallel to 5x−7y=0 and passing through (7, -6) is y = (5/7)x - 36/7.
c) To find the equation of a line perpendicular to 6x+3y=9, we need to find the negative reciprocal of the slope of the given line. Rearranging the equation into slope-intercept form, we get y = -2x + 3. The slope of the given line is -2, so the negative reciprocal is 1/2. Now we can use the point-slope form, y - y1 = m(x - x1), where (x1, y1) is the given point (-2, 3), and m is the slope. Plugging in the values, we get y - 3 = (1/2)(x + 2). Simplifying, we get y = (1/2)x - 1. Therefore, the equation of the line perpendicular to 6x+3y=9 and passing through (-2, 3) is y = (1/2)x - 1.
d) To find the equation of a line perpendicular to 5x−7y=0, we need to find the negative reciprocal of the slope of the given line. Rearranging the equation into slope-intercept form, we get y = (5/7)x. The slope of the given line is (5/7), so the negative reciprocal is -7/5. Now we can use the point-slope form, y - y1 = m(x - x1), where (x1, y1) is the given point (7, -6), and m is the slope. Plugging in the values, we get y - (-6) = (-7/5)(x - 7). Simplifying, we get y = (-7/5)x + 54/5. Therefore, the equation of the line perpendicular to 5x−7y=0 and passing through (7, -6) is y = (-7/5)x + 54/5.