Consider the line -9x+8y=5
Find the equation of the line that is perpendicular to this line and passes through the point (-7,-5)
Find the equation of the line that is parallel to this line and passes through the point l(-7,-5)
First, we need to find the slope of the given line. The slope of a line in the form Ax + By = C is given by -A/B. Therefore, the slope of the line -9x+8y=5 is -(-9)/8 = 9/8.
1. Perpendicular line:
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. Therefore, the slope of the perpendicular line will be -8/9.
Using the point-slope form of a line (y - y1 = m(x - x1)), where m is the slope and (x1, y1) is the given point (-7, -5), we get:
y - (-5) = -8/9(x - (-7))
y + 5 = -8/9(x + 7)
y + 5 = -8/9x - 56/9
y = -8/9x - 56/9 - 45/9
y = -8/9x - 101/9
Therefore, the equation of the line perpendicular to -9x+8y=5 and passing through the point (-7, -5) is y = -8/9x - 101/9.
2. Parallel line:
The slope of a line parallel to another line is the same as the slope of the original line. Therefore, the slope of the parallel line will be 9/8.
Using the point-slope form as before, we get:
y - (-5) = 9/8(x - (-7))
y + 5 = 9/8(x + 7)
y + 5 = 9/8x + 63/8
y = 9/8x + 63/8 - 40/8
y = 9/8x + 23/8
Therefore, the equation of the line parallel to -9x+8y=5 and passing through the point (-7, -5) is y = 9/8x + 23/8.