Write an equation in slope-intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (-2,-6); y=5

Let,the equation of line be
y
=
m
x
+
c
, where,
m
is its slope and
c
is the
Y
intercept.
Now,arranging the given equation in the above mentioned form to get its slope,
Given,
y
+
5
=
3
(
x
−
2
)
=
y
=
3
x
−
11
So,its slope is
3
Now,for two lines to be mutually perpendicular,their product of slope must be
−
1
So,
m
⋅
3
=
−
1
or.
m
=
−
1
3
So,our required line equation becomes,
y
=
−
1
3
x
+
c
Now,given,that the line passes through
(
6
,
2
)
,so putting the values in the equation to get
c
So,
2
=
(
−
1
3
)
⋅
6
+
c
or,
c
=
4
So,the equation of the line is
y
=
−
1
3
x
+
4
or,
3
y
+
x
=
12
graph{3y+x=12 [-10, 10, -5, 5]}