I assume there are some plus signs that aren't rendering for some reason, so that the plane should be

.
You're minimizing

subject to the constraint

. Note that

and

attain their extrema at the same values of

, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.
The Lagrangian is

Take your partial derivatives and set them equal to 0:

Adding the first three equations together yields

and plugging this into the first three equations, you find a critical point at

.
The squared distance is then

, which means the shortest distance must be

.
I just did the second one, the one that looks this a side ways L,
here are the points
(4,4). (5,4). (3,1). (3,2). (4,2). hope this helped
Given:
The sum of two terms of GP is 6 and that of first four terms is 
To find:
The sum of first six terms.
Solution:
We have,


Sum of first n terms of a GP is
...(i)
Putting n=2, we get


...(ii)
Putting n=4, we get



(Using (ii))
Divide both sides by 6.
Taking square root on both sides, we get

Case 1: If r is positive, then using (ii) we get
The sum of first 6 terms is




Case 2: If r is negative, then using (ii) we get
The sum of first 6 terms is




Therefore, the sum of the first six terms is 7.875.
Equations are steeper when their slope's positive value is higher (ignore negatives when determining how steep a slope is). The slope is the value in front of x. If there isn't a value in front of x, then it is assumed to be 1.
I believe the answer is -51+60 because it equals -9. And -51-(-60) is 9.