1. Find the equation of the set of all points P(x,y) that is equidistant from (-3,0) and (3,-5).

2. Find the equation of the set of all points P(x,y) that is twice as far from (-8,8) as from (-2,2).

3. Find the equation of the set of all points P(x,y) such that the sum of the distance from (-4,0) and (4,0) is equal to 12.

Assuming you just want verification of your results,

1. y + 5/2 = 6/5 x

2. (x+8)^2+(y-8)^2 = 4((x+2)^2+(y-2)^2)
or, x^2+y^2 = 32

3. Clearly this is the ellipse with foci at the two points, and a=6, c=4, b^2=20

x^2/36 + y^2/20 = 1

see http://www.wolframalpha.com/input/?i=ellipse+x%5E2%2F36+%2B+y%5E2%2F20+%3D+1

To find the equation of the set of all points P(x,y) in each of these cases, we will use the concept of distance formula and algebraic manipulation. Let's go through each question step by step:

1. Find the equation of the set of all points equidistant from (-3,0) and (3,-5):

To solve this, we need to find the distance between P(x,y) and each of the given points and equate them.

The distance formula is:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Let's use (-3,0) as the first point (x1, y1) and (3,-5) as the second point (x2, y2).

So, we have:
d1 = sqrt((x - (-3))^2 + (y - 0)^2) --- distance from P(x,y) to (-3,0)
d2 = sqrt((x - 3)^2 + (y - (-5))^2) --- distance from P(x,y) to (3,-5)

Since we want P(x,y) to be equidistant from both points, we can equate these distances:
d1 = d2

Now, let's square both sides to eliminate the square root:
d1^2 = d2^2
((x - (-3))^2 + (y - 0)^2) = ((x - 3)^2 + (y - (-5))^2)

Expanding and simplifying:
(x + 3)^2 + y^2 = (x - 3)^2 + (y + 5)^2

Further simplifying, we get:
(x + 3)^2 + y^2 = (x - 3)^2 + (y + 5)^2
x^2 + 6x + 9 + y^2 = x^2 - 6x + 9 + y^2 + 10y + 25

Simplifying further, we have:
6x + 10y = 0

Therefore, the equation of the set of all points P(x,y) equidistant from (-3,0) and (3,-5) is 6x + 10y = 0.

2. Find the equation of the set of all points P(x,y) that is twice as far from (-8,8) as from (-2,2):

Again, we'll use the distance formula to find the distances and equate them.

Let (-8,8) be the first point (x1, y1) and (-2,2) be the second point (x2, y2) in this case.

Using the distance formula:
d1 = sqrt((x - (-8))^2 + (y - 8)^2) --- distance from P(x,y) to (-8,8)
d2 = sqrt((x - (-2))^2 + (y - 2)^2) --- distance from P(x,y) to (-2,2)

Since we want P(x,y) to be twice as far from (-8,8) as from (-2,2), we can write:
d1 = 2 * d2

Now, square both sides:
d1^2 = (2 * d2)^2
((x - (-8))^2 + (y - 8)^2) = 4 * ((x - (-2))^2 + (y - 2)^2)

Expanding and simplifying:
(x + 8)^2 + (y - 8)^2 = 4 * (x + 2)^2 + 4 * (y - 2)^2

Further simplifying, we get:
x^2 + 16x + 64 + y^2 - 16y + 64 = 4x^2 + 16x + 16 + 4y^2 - 16y + 16

Simplifying further, we have:
3x^2 + 3y^2 - 48 = 0

Therefore, the equation of the set of all points P(x,y) that is twice as far from (-8,8) as from (-2,2) is 3x^2 + 3y^2 - 48 = 0.

3. Find the equation of the set of all points P(x,y) such that the sum of the distance from (-4,0) and (4,0) is equal to 12:

Again, we'll use the distance formula to find the distances and equate their sum to 12.

Using (-4,0) as the first point (x1, y1) and (4,0) as the second point (x2, y2):

d1 + d2 = 12

d1 = sqrt((x - (-4))^2 + (y - 0)^2) --- distance from P(x,y) to (-4,0)
d2 = sqrt((x - 4)^2 + (y - 0)^2) --- distance from P(x,y) to (4,0)

Therefore:
sqrt((x + 4)^2 + y^2) + sqrt((x - 4)^2 + y^2) = 12

To eliminate the square root, we can square both sides:
((x + 4)^2 + y^2) + 2 * sqrt((x + 4)^2 + y^2) * sqrt((x - 4)^2 + y^2) + ((x - 4)^2 + y^2) = 144

Simplifying, we get:
2x^2 + 2y^2 - 96 = 0

Therefore, the equation of the set of all points P(x,y) such that the sum of the distance from (-4,0) and (4,0) is equal to 12 is 2x^2 + 2y^2 - 96 = 0.