Write an equation in standard form of the line that contains the point (-3,4) and is a. parallel to the line y=7x+12 and b. perpendicular to the line y=7x+12
a. To find the equation of the line that is parallel to y = 7x + 12, we know that parallel lines have the same slope. The given line has a slope of 7. Using the point-slope form of a linear equation, y - y₁ = m(x - x₁), where (x₁, y₁) is the given point, we can substitute the given point (-3, 4) and the slope 7 into the equation to find the equation of the line:
y - 4 = 7(x + 3)
Expanding the equation:
y - 4 = 7x + 21
Rearranging the equation in standard form:
7x - y = -25
Therefore, the equation of the line parallel to y = 7x + 12 and containing the point (-3, 4) is 7x - y = -25.
b. To find the equation of the line that is perpendicular to y = 7x + 12, we know that perpendicular lines have slopes that are negative reciprocals. The given line has a slope of 7, so the perpendicular line will have a slope of -1/7. Using the point-slope form of a linear equation, we can substitute the given point (-3, 4) and the slope -1/7 into the equation to find the equation of the line:
y - 4 = (-1/7)(x + 3)
Expanding the equation:
y - 4 = -x/7 - 3/7
Adding x/7 to both sides:
x/7 + y - 4 = -3/7
Rearranging the equation in standard form by multiplying through by 7:
x + 7y - 28 = -3
Adding 3 to both sides:
x + 7y - 25 = 0
Therefore, the equation of the line perpendicular to y = 7x + 12 and containing the point (-3, 4) is x + 7y - 25 = 0.