Question 115438
A line parallel to y = -(1/2)x + 2 is of the form, y = -(1/2)x + b where b is an arbitrary constant.

Now, the required line passes through (3,3).

so, (3,3)will satify y = -(1/2)x + b.

Substitute (3,3) in y = -(1/2)x + b.

We get, 3 = -(1/2)3 + b.

That is, 3 = -(3/2)+ b

Or, b = 3 + 3/2

Or, b = 9/2

So, the equation of the required line is, y = -(1/2)x + (9/2)

Or, 2y = -x + 9.