If Q (0, 1) is equidistant from P (5,-3) and R (x, 6) , then positive value of x is

Options

A. 5
B. 4
C. 2
D. 8

Answer

To find the positive value of \( x \) such that point \( Q(0, 1) \) is equidistant from points \( P(5, -3) \) and \( R(x, 6) \), we can follow these steps:

### Step 1: Calculate the distance between points \( Q \) and \( P \)

The distance \( d(Q, P) \) can be calculated using the distance formula:

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

Substituting the coordinates of points \( Q(0, 1) \) and \( P(5, -3) \):

\[ d(Q, P) = \sqrt{(5 - 0)^2 + (-3 - 1)^2} \]
\[ = \sqrt{(5)^2 + (-4)^2} \]
\[ = \sqrt{25 + 16} \]
\[ = \sqrt{41} \]

### Step 2: Calculate the distance between points \( Q \) and \( R \)

Similarly, we calculate the distance \( d(Q, R) \):

\[ d(Q, R) = \sqrt{(x - 0)^2 + (6 - 1)^2} \]
\[ = \sqrt{x^2 + (5)^2} \]
\[ = \sqrt{x^2 + 25} \]

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

Since \( Q \) is equidistant from \( P \) and \( R \), we set the two distances equal:

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

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

Squaring both sides gives:

\[ 41 = x^2 + 25 \]

### Step 5: Solve for \( x^2 \)

Rearranging the equation:

\[ x^2 = 41 - 25 \]
\[ x^2 = 16 \]

### Step 6: Find the positive value of \( x \)

Taking the square root of both sides:

\[ x = \sqrt{16} \]
\[ x = 4 \]

Thus, the positive value of \( x \) is \( 4 \).

### Final Answer:
The positive value of \( x \) is \( 4 \).

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