Find the slope-intercept equation of the line with the following properties. Perpendicular to the line -9x-8y=96; containing the point (-8,-8)
Can anyone please explain and help me please.
slope of given line: -8/9
slope of perpendicular is 9/8
so, using the point-slope form,
(y+8) = 9/8 (x+8)
To find the slope-intercept equation of a line that is perpendicular to another line, you need to determine the slopes of both lines.
First, let's rewrite the given equation -9x - 8y = 96 in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
Starting with -9x - 8y = 96:
1. Subtract -9x from both sides: -8y = 9x + 96
2. Divide both sides by -8 to isolate y: y = -9/8x - 12
So the given line has a slope of -9/8.
Since the line we want to find is perpendicular to the given line, it will have a slope that is the negative reciprocal of -9/8. The negative reciprocal of a fraction is obtained by flipping the fraction and changing the sign.
Thus, the slope of the line we are looking for is 8/9.
Now we have the slope of the line and a point (-8,-8) that it passes through. We can use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Using the point-slope form and substituting the given values, we have:
y - (-8) = (8/9)(x - (-8))
Simplifying this equation:
y + 8 = 8/9(x + 8)
y + 8 = (8/9)x + (64/9)
Rearranging the equation to slope-intercept form (y = mx + b):
y = (8/9)x + (64/9) - (72/9)
y = (8/9)x - 8/9
Therefore, the slope-intercept equation of the line that is perpendicular to -9x - 8y = 96 and passes through the point (-8,-8) is y = (8/9)x - 8/9.