What is the equation of a line that is perpendicular to y=3x-2 and passes through the point (6 8) ?
a)y=-1/3x+10
b)y=1/3x+6
c)y=-1/3x+4
d)y=-1/3x+8

The result is a line perpendicular to y=3x-2
This line has the following equation: y =mx + b, where m = slope

Remember that the product of the slopes of two perpendicular lines is always = -1 (or in other term one is the reciprocal inverse of the other) so the
first slope = 3 and the sope perpendicular to 3 will be - 1/3.
Then the new equation is y = - 1/3(x) + b
How to calculate b? This line passes through (6, 8), that
means (x=6 and y=8) . Plug these values in y = -1/3 (x)+b:3
 8=(-1/3)(6) + b; 
8= - 2 + b and b = 10
Te final equation is y = - 1/3.x+10 (answer A)

A: 0.26 would be the answer