Find the equation of a line that is perpendicular to the line y=1/8x+5 and contains the points (-7,0)
y=
please show work
slope m = - 1/(1/8) = -8
y = m x + b
0 = -8(-7) + b
- 56 = b
so
y = -8 x - 56
To find the equation of a line that is perpendicular to the given line and contains the point (-7, 0), we need to determine the slope of the perpendicular line first.
The given line has a slope of 1/8 because it is in the form y = mx + b, where m represents the slope.
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. So, the slope of the line we are looking for is -8/1 or -8.
Now we have the slope of the perpendicular line (-8) and one point that the line passes through (-7, 0).
We can use the point-slope formula to find the equation of the line:
y - y1 = m(x - x1)
Let's plug in the values:
y - 0 = -8(x - (-7))
Simplifying:
y = -8(x + 7)
Expanding, we get:
y = -8x - 56
Therefore, the equation of the line that is perpendicular to y = 1/8x + 5 and contains the point (-7, 0) is y = -8x - 56.