To find the equation of the line perpendicular to y = 7/(3x + 6) and passing through (-6,-4), we first need to find the slope of the original line. The slope of y = 7/(3x + 6) is the coefficient of x, which is 7/3.
The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line. Therefore, the slope of the perpendicular line is -3/7.
Using the point-slope form of the equation of a line, the equation of the line perpendicular to y = 7/(3x + 6) and passing through (-6,-4) can be written as:
y - y1 = m(x - x1)
y - (-4) = -3/7(x - (-6))
y + 4 = -3/7(x + 6)
y + 4 = -3/7x - 18/7
y = -3/7x - 46/7
Therefore, the equation of the line perpendicular to y = 7/(3x + 6) and passing through (-6,-4) is y = -3/7x - 46/7.
To find the equation of the line parallel to y = 7/(3x + 6) and passing through (-6,-4), we again use the slope of the original line, which is 7/3.
The slope of a line parallel to a given line is equal to the slope of the given line. Therefore, the slope of the line parallel to y = 7/(3x + 6) is 7/3.
Using the point-slope form of the equation of a line, the equation of the line parallel to y = 7/(3x + 6) and passing through (-6,-4) can be written as:
y - y1 = m(x - x1)
y - (-4) = 7/3(x - (-6))
y + 4 = 7/3(x + 6)
y + 4 = 7/3x + 14
y = 7/3x + 10
Therefore, the equation of the line parallel to y = 7/(3x + 6) and passing through (-6,-4) is y = 7/3x + 10.
Consider the line y= 7 divided by 3x +6
Find the equation of the line that is perpendicular to this line and passes through the point (-6,-4).
Find the equation of the line that is parallel to this line and passes through the point l(-6,-4)
1 answer
1 answer