Given: Vertices of the parallelogram are A(-2, -1), B(1, 0), C(x, 3) and D(1, y).

To find: values of x and y.

Since, ABCD is a parallelogram, we have AB = CD and BC = DA.

AB

= √10 units

BC

CD

DA

Since AB = CD,

Squaring both sides, we get

⇒ 10 = (1 – x)² + (y – 3)²

⇒ 10 = 1 – 2x + x² + y² – 6y + 9

⇒ x² + y² – 2x – 6y = 0 …..(1)

Since BC = DA,

Squaring both sides,

⇒ (x – 1)² + 9 = 9 + (1 + y)²

⇒ x² – 2x + 1 = 1 + 2y + y²

⇒ x² – y² – 2x – 2y = 0 …..(2)

Equation 1 – Equation 2 gives us,

⇒ 2y² – 4y = 0

⇒ y² – 2y = 0

⇒ y(y – 2) = 0

⇒ y = 0 or y = 2

But y 0 because then point D(1, 0) is same as B(1, 0)

Therefore, y = 2

When y = 2, from equation 1,

⇒ x² + 4 – 2x – 12 = 0

⇒ x² – 2x – 8 = 0

⇒ (x – 4) × (x + 2) = 0

⇒ x = 4 or x = -2

So, the possible set of values for x and y are:

x = 4, y = 2

x = -2, y = 2

But when x = -2, then C(-2, 3). Then ABCD does not form a parallelogram.

Therefore, the only solution is x = 4 and y = 2.