If the point A (2, -4) is equidistant from P (3, 8) and Q (-10, y), find the values of y. Also find distance PQ.

Answer

To solve the problem, we need to find the value of \( y \) such that the distance from point \( A(2, -4) \) to point \( P(3, 8) \) is equal to the distance from point \( A(2, -4) \) to point \( Q(-10, y) \). We will also find the distance \( PQ \).

### Step 1: Calculate the distance \( AP \)

The distance 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 points \( A(2, -4) \) and \( P(3, 8) \):

\[ AP = \sqrt{(3 - 2)^2 + (8 - (-4))^2} \]
\[ = \sqrt{(1)^2 + (8 + 4)^2} \]
\[ = \sqrt{1 + 12^2} \]
\[ = \sqrt{1 + 144} \]
\[ = \sqrt{145} \]

### Step 2: Calculate the distance \( AQ \)

Now, we calculate the distance \( AQ \) where \( Q(-10, y) \):

\[ AQ = \sqrt{(-10 - 2)^2 + (y - (-4))^2} \]
\[ = \sqrt{(-12)^2 + (y + 4)^2} \]
\[ = \sqrt{144 + (y + 4)^2} \]

### Step 3: Set the distances equal

Since \( A \) is equidistant from \( P \) and \( Q \), we set \( AP = AQ \):

\[ \sqrt{145} = \sqrt{144 + (y + 4)^2} \]

### Step 4: Square both sides to eliminate the square roots

\[ 145 = 144 + (y + 4)^2 \]

### Step 5: Simplify the equation

\[ 145 - 144 = (y + 4)^2 \]
\[ 1 = (y + 4)^2 \]

### Step 6: Solve for \( y \)

Taking the square root of both sides:

\[ y + 4 = 1 \quad \text{or} \quad y + 4 = -1 \]

1. For \( y + 4 = 1 \):
\[ y = 1 - 4 = -3 \]

2. For \( y + 4 = -1 \):
\[ y = -1 - 4 = -5 \]

Thus, the possible values of \( y \) are \( -3 \) and \( -5 \).

### Step 7: Calculate the distance \( PQ \)

Now we will find the distance \( PQ \) using the coordinates of \( P(3, 8) \) and \( Q(-10, y) \):

Using \( y = -3 \):
\[ PQ = \sqrt{(-10 - 3)^2 + (-3 - 8)^2} \]
\[ = \sqrt{(-13)^2 + (-11)^2} \]
\[ = \sqrt{169 + 121} \]
\[ = \sqrt{290} \]

Using \( y = -5 \):
\[ PQ = \sqrt{(-10 - 3)^2 + (-5 - 8)^2} \]
\[ = \sqrt{(-13)^2 + (-13)^2} \]
\[ = \sqrt{169 + 169} \]
\[ = \sqrt{338} \]

### Final Answers

The values of \( y \) are \( -3 \) and \( -5 \). The distances \( PQ \) are \( \sqrt{290} \) when \( y = -3 \) and \( \sqrt{338} \) when \( y = -5 \).