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Ans) S = {(x, y) ∈ R^2: x + 2y = 4}

Let U, V ∈ S, where U = (u1,u2) and V = (v1,v2)

Then u1+ 2u2 = 4 and v1 + 2v2 = 4 -----(i)

Let W = (w1, w2) is a point on the line segment joining the points U and V, then

W = λU + (1 – λ)V, 0 ≤ λ ≤ 1

Hence, (w1, w2) = λ(u1, u2) + (1 – λ) (v1, v2)

Now w1 + 2w2 = [λu1 + (1 – λ) v1] + 2[λu2 + (1 – λ) v2]

= λ(u1 + 2u2) + (1- λ)(v1 + 2v2)

= 4 λ + 4(1 – λ) = 4 Using equation (i)

Thus, W = w1, w2 ∈ S

Hence, S is a convex set.

Let U, V ∈ S, where U = (u1,u2) and V = (v1,v2)

Then u1+ 2u2 = 4 and v1 + 2v2 = 4 -----(i)

Let W = (w1, w2) is a point on the line segment joining the points U and V, then

W = λU + (1 – λ)V, 0 ≤ λ ≤ 1

Hence, (w1, w2) = λ(u1, u2) + (1 – λ) (v1, v2)

Now w1 + 2w2 = [λu1 + (1 – λ) v1] + 2[λu2 + (1 – λ) v2]

= λ(u1 + 2u2) + (1- λ)(v1 + 2v2)

= 4 λ + 4(1 – λ) = 4 Using equation (i)

Thus, W = w1, w2 ∈ S

Hence, S is a convex set.

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