Thread: First degree polynomial function f(x,y)

So Im trying to find a function of a plane with:
n = <12,-3,3> and passes through the point (2,5-2)

and it needs to be in terms of f(x,y)
I don't know if this is right but this is what I did...

12(x-2) - 3(y-5) + 3(z+2) = 0
...
12x - 3y +3z = 3
...
z = 4x - y -1

so therefore, isn't z another way of saying f(x,y) ?
so then f(x,y) = 4x - y - 1?

Originally Posted by RET80

So Im trying to find a function of a plane with:
n = <12,-3,3> and passes through the point (2,5-2)

and it needs to be in terms of f(x,y)
I don't know if this is right but this is what I did...

12(x-2) - 3(y-5) + 3(z+2) = 0
...
12x - 3y +3z = 3 3z = 3 -12x +3y z = 1 -4x + y
...
z = 4x - y -1

so therefore, isn't z another way of saying f(x,y) ?
so then f(x,y) = 4x - y - 1?

Thus,