Find the values of $$ y $$ such that the distance between the points $$ (-1,1) $$ and $$ (3, y) $$ is 5 units. The values of $$ y $$ are $$ \square $$ and $$ \square $$.

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To find the values of $$ y $$ such that the distance between the points $$ (-1, 1) $$ and $$ (3, y) $$ is 5 units, we can use the distance formula. The distance $$ d $$ between two points $$ (x_1, y_1) $$ and $$ (x_2, y_2) $$ is given by:

$$
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

In our case, the points are $$ (-1, 1) $$ and $$ (3, y) $$. Thus, we have:

- $$ x_1 = -1 $$
- $$ y_1 = 1 $$
- $$ x_2 = 3 $$
- $$ y_2 = y $$

Substituting these values into the distance formula, we set the distance equal to 5:

$$
\sqrt{(3 - (-1))^2 + (y - 1)^2} = 5
$$

This simplifies to:

$$
\sqrt{(3 + 1)^2 + (y - 1)^2} = 5
$$

$$
\sqrt{4^2 + (y - 1)^2} = 5
$$

$$
\sqrt{16 + (y - 1)^2} = 5
$$

Next, we square both sides to eliminate the square root:

$$
16 + (y - 1)^2 = 25
$$

Now, we isolate $$ (y - 1)^2 $$:

$$
(y - 1)^2 = 25 - 16
$$

$$
(y - 1)^2 = 9
$$

Taking the square root of both sides gives us two equations:

$$
y - 1 = 3 \quad \text{or} \quad y - 1 = -3
$$

Solving these equations:

1. For $$ y - 1 = 3 $$:

$$
y = 3 + 1 = 4
$$

2. For $$ y - 1 = -3 $$:

$$
y = -3 + 1 = -2
$$

Thus, the values of $$ y $$ such that the distance between the points is 5 units are:

$$
\boxed{4} \quad \text{and} \quad \boxed{-2}
$$
The values of $$ y $$ are 4 and -2.

Find the values of such that the distance between the points and is 5 units.
The values of are and .

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Find the values of such that the distance between the points and is 5 units. The values of are and .