Let A(3, -1) and B(1, 1) be the end points of line segment AB. Let P be the middle point of the line segment AB. Let Q be the point situated at a distance \(\sqrt{2}\) units from P on the perpendicular bisector line of AB. What are the possible coordinates of Q?

Calculation:

Given points A(3, −1) and B(1, 1). Let P be the midpoint of AB, and Q be a point on the perpendicular bisector of AB that lies √2 units from P.

Compute the midpoint P of AB:

\(P = \Bigl(\tfrac{3 + 1}{2},\,\tfrac{-1 + 1}{2}\Bigr) = (2,\,0) \)

Compute the slope of AB:

\(m_{AB} = \frac{\,1 - (-1)\,}{\,1 - 3\,} = \frac{2}{-2} = -1\)

Therefore, the equation of line AB is:

\(y - 1 = -1\,(x - 1)\;\Longrightarrow\;x + y - 2 = 0\)

The perpendicular bisector of AB must pass through P(2, 0) and have slope perpendicular to −1 (i.e. slope +1):

\(y - 0 = 1\,(x - 2)\;\Longrightarrow\;y = x - 2\)

Any point Q on this bisector satisfies y = x - 2. Write Q = (x, x − 2).

We require the distance PQ = √2. Since P(2, 0),

\(\text{Distance}^2 = (x - 2)^2 + \bigl((\,x - 2\,) - 0\bigr)^2 = 2\)

⇒ \((x - 2)^2 + (x - 2)^2 = 2\;\Longrightarrow\;2\,(x - 2)^2 = 2\;\Longrightarrow\;(x - 2)^2 = 1\)

Thus,

⇒ \(x - 2 = \pm 1\;\Longrightarrow\;x = 3\text{ or }x = 1\)

If x = 3, then y = 3 - 2 = 1. So one solution is Q(3, 1).

If x = 1, then y = 1 - 2 = -1. So the other solution is Q(1, −1).

Hence, the correct answer is Option 2.