What is the equation of a line that passes through the point (4, 2) and is perpendicular to the line whose equation is y=x/3−1 ?

The line y = x/3 - 1, or y = (x/3) - 1, is the same as y = (1/3)x - 1
The slope here is 1/3 as this equation is in y = mx+b form (m = slope, b = y intercept)

Using the starting point of 1/3, we will...
Flip the fraction: 1/3 ----> 3/1 = 3
Then flip the sign from positive to negative:  3 ----> -3

The original slope is 1/3
The perpendicular slope is -3
The two slopes multiply to -1
We'll use m = -3 along with x = 4 and y = 2
We generate those x,y values from (x,y) = (4,2) as the given point

Plug m = -3, x = 4, y = 2 into the equation below. Solve for b
y = mx+b
2 = -3*4 + b
2 = -12 + b
2+12 = -12+b+12
14 = b
b = 14

So because m = -3 and b = 14, we go from y = mx+b to y = -3x+14

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Final y = -3x+14

Visual proof can be found using software like GeoGebra as shown in the attached image.

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