If Q (0, 1) is equidistant from P(5, -3) and R (x, 6), find the values of 'x'. Also, find the distances of QR and PR.

Answer

To solve the problem, we need to find the value of 'x' such that the point Q(0, 1) is equidistant from the points P(5, -3) and R(x, 6). We will also calculate the distances QR and PR.

### Step 1: Calculate the distance PQ

Using the distance formula between two points (x1, y1) and (x2, y2):

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

For points P(5, -3) and Q(0, 1):

\[ PQ = \sqrt{(5 - 0)^2 + (-3 - 1)^2} \]

Calculating this:

\[ PQ = \sqrt{(5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} \]

### Step 2: Calculate the distance QR

For points Q(0, 1) and R(x, 6):

\[ QR = \sqrt{(x - 0)^2 + (6 - 1)^2} \]

Calculating this:

\[ QR = \sqrt{x^2 + (5)^2} = \sqrt{x^2 + 25} \]

### Step 3: Set the distances equal to each other

Since Q is equidistant from P and R, we set the distances equal:

\[ PQ = QR \]

Substituting the distances we found:

\[ \sqrt{41} = \sqrt{x^2 + 25} \]

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

\[ 41 = x^2 + 25 \]

### Step 5: Solve for x

Subtract 25 from both sides:

\[ 41 - 25 = x^2 \]

\[ 16 = x^2 \]

Taking the square root of both sides gives:

\[ x = 4 \quad \text{or} \quad x = -4 \]

### Step 6: Calculate the distances QR and PR

1. **For R(4, 6)**:
- Calculate QR:
\[ QR = \sqrt{(4 - 0)^2 + (6 - 1)^2} = \sqrt{16 + 25} = \sqrt{41} \]

- Calculate PR:
\[ PR = \sqrt{(5 - 4)^2 + (-3 - 6)^2} = \sqrt{1 + 81} = \sqrt{82} \]

2. **For R(-4, 6)**:
- Calculate QR:
\[ QR = \sqrt{(-4 - 0)^2 + (6 - 1)^2} = \sqrt{16 + 25} = \sqrt{41} \]

- Calculate PR:
\[ PR = \sqrt{(5 - (-4))^2 + (-3 - 6)^2} = \sqrt{(5 + 4)^2 + (-9)^2} = \sqrt{81 + 81} = \sqrt{162} \]

### Summary of Results

- The values of 'x' are 4 and -4.
- The distances are:
- For R(4, 6): QR = √41, PR = √82
- For R(-4, 6): QR = √41, PR = √162