What is the equation in slope-intercept form of the line that crosses the x-axis at 12 and is perpendicular to the line represented by y = 6/5x + 9 ?

The equation of the line in slope-intercept form is;

y = 10 - 5/6·x

Step-by-step explanation:

The given equation of the line to which the required line is perpendicular is presented as follows;

y = 6/5·x + 9

A point on the desired line = (12, 0)  

The relationship of the slope, m₁ of a line, perpendicular to the another line of slope m₂, is given as follows;

m₁  = -1/m₂

Therefore;

The slope of the desired line = -1/(The slope of the given line)

The slope of the desired line = -1/(6/5) = -5/6

The equation of the desired line in point and slope form is given as follows;

y - 0 = -5/6 × (x - 12)

Therefore, the equation of the line in slope-intercept form is;

y = -5/6·x + 10 = 10 - 5/6·x.