Given:
P=(6,2)
y=(-1/3)x+6 ..... [line 1]
[Line 2](eqn unknown) is perpendicular to [Line 1]
We know that if two lines are perpendicular,the product of their slopes is -1.
ie. m1*m2=-1
Let slope of line 1 be m1.
Then,
-1/3*m2=-1
m2=(-1*3)/-1
=3
Then slope of required line is 3.
We have one point P(6,2) on this line of slope 3.
Then using point-slope form,
y-y1=m(x-x1)
y-2=3(x-6)
y-2=3x-18
y=3x-18+2
y=3x-16
Equation of the line in standard form: y = 3x - 16
Hope this helps,
best of luck.