A line segment begins at $(2, 5)$. It is 10 units long and ends at the point $(-6, y)$ where $y > 0$. What is the value of $y$?
We use the distance formula: \begin{align*}
\sqrt{(2 - (-6))^2 + (y - 5)^2} &= \sqrt{8^2 + (y - 5)^2} \\
& = \sqrt{y^2 - 10y + 89} \\
& = 10.
\end{align*}Squaring both sides and rearranging terms, we find that \begin{align*}
y^2 - 10y + 89 &= 100 \\
y^2 - 10y - 11 &= 0\\
(y - 11)(y + 1) &= 0
\end{align*}Thus, $y = 11$ or $y = -1$. We are given that $y > 0$, so $y = \boxed{11}$.
11
Level 4
Algebra
Problem: A line segment begins at $(2, 5)$. It is 10 units long and ends at the point $(-6, y)$ where $y > 0$. What is the value of $y$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of \( y \), we start by determining the distance between the two points \( (2, 5) \) and \( (-6, y) \). The distance formula is given by:

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

In this case, \( d = 10 \), \( (x_1, y_1) = (2, 5) \), and \( (x_2, y_2) = (-6, y) \). Substituting these values into the distance formula, we have:

\[
10 = \sqrt{((-6) - 2)^2 + (y - 5)^2}
\]

Calculating the difference in the x-coordinates:

\[
-6 - 2 = -8
\]

Thus, we can rewrite the equation as:

\[
10 = \sqrt{(-8)^2 + (y - 5)^2}
\]

Calculating \( (-8)^2 \):

\[
(-8)^2 = 64
\]

Now substituting this back into the equation gives:

\[
10 = \sqrt{64 + (y - 5)^2}
\]

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

\[
10^2 = 64 + (y - 5)^2
\]

This simplifies to:

\[
100 = 64 + (y - 5)^2
\]

Subtracting 64 from both sides results in:

\[
100 - 64 = (y - 5)^2
\]

This simplifies to:

\[
36 = (y - 5)^2
\]

Taking the square root of both sides, we find:

\[
y - 5 = 6 \quad \text{or} \quad y - 5 = -6
\]

Solving these equations gives:

1. \( y - 5 = 6 \) leads to \( y = 11 \)
2. \( y - 5 = -6 \) leads to \( y = -1 \)

Since the problem states that \( y > 0 \), we discard \( y = -1 \). Therefore, the only valid solution is:

\[
y = 11
\]<|END_OF_LATENT|> The final answer is $11$.