Question

The point on the line \( 4x - y - 2 = 0 \) which is equidistant from the points \( (-5, 6) \) and \( (3, 2) \) is

• A. \( (2, 6) \)
• B. \( (4, 14) \)
• C. \( (1, 2) \)
• D. \( (3, 10) \)

Hint:

To find equidistant points from two given locations, solve the system of equations formed by the distance formula and any other relevant conditions, like the line equation.

Answer 1

The Correct Option is B

We are given the line equation: \[ 4x - y - 2 = 0. \] Since point \( P \) is equidistant from \( A(-5, 6) \) and \( B(3, 2) \), the condition \( PA = PB \) must hold. Therefore, we use the equation: \[ PA^2 = PB^2. \] Step 1: Use the distance formula. The distances from \( P(a, b) \) to \( A(-5, 6) \) and \( B(3, 2) \) are: \[ \sqrt{(a + 5)^2 + (b - 6)^2} = \sqrt{(a - 3)^2 + (b - 2)^2}. \] Squaring both sides, we get: \[ (a + 5)^2 + (b - 6)^2 = (a - 3)^2 + (b - 2)^2. \] Simplifying: \[ a^2 + 10a + 25 + b^2 - 12b + 36 = a^2 - 6a + 9 + b^2 - 4b + 4. \] Combine like terms: \[ 16a - 8b + 48 = 0. \] Step 2: Solve the system of equations. From the line equation \( 4x - y - 2 = 0 \), we can express \( b \) in terms of \( a \) as: \[ b = 4a - 2. \] Substitute this into the simplified distance equation: \[ 16a - 8(4a - 2) + 48 = 0. \] Simplify: \[ 16a - 32a + 16 + 48 = 0, \] \[ -16a + 64 = 0. \] Solving for \( a \): \[ a = 4. \] Substitute \( a = 4 \) into the equation for \( b \): \[ b = 4(4) - 2 = 14. \] Final Answer: The coordinates of point \( P \) are: \[ \boxed{(4, 14)}. \]