Find the values of y for which the distance between the points `P(2,\ -3)` and `Q(10 ,\ y)`is 10 units.

Options

A. `y=9`
B. `y=-5`
C. `y=-3`
D. Both A and C

Explanation

To find the values of \( y \) for which the distance between the points \( P(2, -3) \) and \( Q(10, y) \) is 10 units, we can follow these steps:

### Step 1: Use the Distance Formula
The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For our points \( P(2, -3) \) and \( Q(10, y) \), we have:
- \( x_1 = 2 \)
- \( y_1 = -3 \)
- \( x_2 = 10 \)
- \( y_2 = y \)

### Step 2: Set Up the Equation
We know the distance \( d \) is 10 units, so we set up the equation:
\[ 10 = \sqrt{(10 - 2)^2 + (y - (-3))^2} \]

### Step 3: Simplify the Equation
Calculating \( (10 - 2)^2 \):
\[ 10 - 2 = 8 \quad \Rightarrow \quad (10 - 2)^2 = 8^2 = 64 \]
So, the equation becomes:
\[ 10 = \sqrt{64 + (y + 3)^2} \]

### Step 4: Square Both Sides
To eliminate the square root, we square both sides:
\[ 10^2 = 64 + (y + 3)^2 \]
This simplifies to:
\[ 100 = 64 + (y + 3)^2 \]

### Step 5: Isolate the Square Term
Next, isolate \( (y + 3)^2 \):
\[ 100 - 64 = (y + 3)^2 \]
\[ 36 = (y + 3)^2 \]

### Step 6: Take the Square Root
Taking the square root of both sides gives us two equations:
\[ y + 3 = 6 \quad \text{or} \quad y + 3 = -6 \]

### Step 7: Solve for \( y \)
Now, we solve for \( y \) in both cases:
1. \( y + 3 = 6 \)
\[ y = 6 - 3 = 3 \]
2. \( y + 3 = -6 \)
\[ y = -6 - 3 = -9 \]

### Final Answer
The values of \( y \) for which the distance between the points \( P(2, -3) \) and \( Q(10, y) \) is 10 units are:
\[ y = 3 \quad \text{and} \quad y = -9 \]