Write an equation of the form y = kx + b, the graph of which passes through the points A (-3; 3) and P (3; -3).

The condition for point A (-3; 3) to belong to the straight line y = kx + b will be:
-3 * k + b = 3, and points P (3; -3):
3 * k + b = -3.
This system is most easily solved by the subtraction method: subtract the second from the first equation (b mutually cancel):
-6 * k = 6
k = -1
Substitute in the first equation and find b:
-3 * (- 1) + b = 3
b = 3 – 3
b = 0
y = -x