Answer:
x+3y-6=0
Step-by-step explanation:
given eqn is y=3x-2 which is 3x-y-2=0
the eqn of line perpendicular to given eqn is -x+3y+k=0
it passes through (6,4)
-6+3*4+k=0
or,. -6+12+k=0
or, k= -6
therefore, the eqn of line perpendicular to given eqn is x+3y-6=0
<span> 100 m / 50 s = 2 m/s = 2 *3600 / 1000 = 7.2 k/h
hope it helps :)</span>
12/100 = 3/25
hope it helps