Part A)
the correct question is
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
we have that
y = (x/3)- 1
step 1
find the slope m
m=(1/3)
we know that
if two lines are perpendicular
so
the slopes
m1*m2=-1
then
(1/3)*m2=-1
m2=-3
step 2
with m=-3 and the point (4,2) find the equation of the line
y-y1=m*(x-x1)------> y-2=-3*(x-4)----> y-2=-3x+12----> y=-3x+14
the answer part A) is
y=-3x+14
Part B)
the correct question is
What is the equation of a line that passes through the point (4,2) and is parallel to the line whose equation is y = (x/3)- 1
we have that
y = (x/3)- 1
step 1
find the slope m
m=(1/3)
we know that
if two lines are parallel
so
the slopes
m1=m2
then
m2=(1/3)
step 2
with m=(1/3) and the point (4,2) find the equation of the line
y-y1=m*(x-x1)------> y-2=(1/3)*(x-4)----> y-2=(1/3)x-(4/3)
y=(1/3)x-(4/3)+2-----> y=(1/3)x+(2/3)
the answer part B) is
y=(1/3)x+(2/3)
using a graph tool
see the attached figure