find the equation of a line that is perpendicular to line y=1/6x+6 and contains point (-7,0)
y=?
please show work so i can see how its done
slope of y = (1/6)x + 6 is 1/6
so slope of perpendicular is -6
equation is y = -6x + b
but (-7,0) lies on it
0 = -6(-7) + b
b = -42
y = -6x - 42
To find the equation of a line that is perpendicular to the line y = (1/6)x + 6, we first need to determine the slope of the perpendicular line.
The given line is in slope-intercept form, y = mx + b, where m represents the slope.
Comparing the given equation to the slope-intercept form, we can see that the slope of the given line is 1/6.
Since the two lines are perpendicular, the slope of the perpendicular line will be the negative reciprocal of 1/6.
The negative reciprocal is obtained by flipping the fraction and changing its sign. Therefore, the slope of the perpendicular line is -6/1, which can be simplified to -6.
Now that we have the slope of the perpendicular line (-6) and a point it passes through (-7, 0), we can use the point-slope form of a linear equation to find the equation of the line.
The point-slope form of a linear equation is given by y - y1 = m(x - x1), where (x1, y1) represents the given point and m is the slope.
Plugging in the values:
y - 0 = -6(x - (-7))
Simplifying the equation:
y = -6(x + 7)
Expanding the equation:
y = -6x - 42
Therefore, the equation of the line that is perpendicular to y = (1/6)x + 6 and contains the point (-7, 0) is y = -6x - 42.