Consider the line y=7x-4
Find the equation of the line that is perpendicular to this line and passes through the point (-8,6)
Find the equation of the line that is parallel to this line and passes through the point l(-8,6)
To find the equation of the line that is perpendicular to y=7x-4 and passes through the point (-8,6), we start by determining the slope of the given line. The slope of the line y=7x-4 is 7.
For a line perpendicular to this line, the slope will be the negative reciprocal of 7, which is -1/7. So the slope of the perpendicular line is -1/7.
Using the point-slope formula for a line, we plug in the slope m=-1/7 and the point (-8,6) to find the equation of the perpendicular line:
y - y1 = m(x - x1)
y - 6 = -1/7(x + 8)
y - 6 = -1/7x - 8/7
y = -1/7x + 34/7
Therefore, the equation of the line that is perpendicular to y=7x-4 and passes through the point (-8,6) is y = -1/7x + 34/7.
To find the equation of the line that is parallel to y=7x-4 and passes through the point (-8,6), we know that parallel lines have the same slope. Therefore, the slope of the parallel line is also 7.
Using the point-slope formula again, we plug in the slope m=7 and the point (-8,6) to find the equation of the parallel line:
y - y1 = m(x - x1)
y - 6 = 7(x + 8)
y - 6 = 7x + 56
y = 7x + 62
Therefore, the equation of the line that is parallel to y=7x-4 and passes through the point (-8,6) is y = 7x + 62.