Let's solve the problem in **2 steps** with explanations at each step: --- ### **Step 1: Describe the Set \( S_1 \) and Its Elements** The solution set \( S_1 \) of the equation \( y = kx \) is: \[ S_1 = \{(x, y) \in \mathbb{R}^2 : y = kx\} \] This means every element in \( S_1 \) can be written as \( (x, kx) \) where \( x \in \mathbb{R} \). **Explanation:** This is a line through the origin in \( \mathbb{R}^2 \) with slope \( k \). --- ### **Step 2: Verify the Vector Space Axioms** Let's check the important vector space properties for \( S_1 \): #### **Closure under Addition** Take any two elements \( (x_1, kx_1) \) and \( (x_2, kx_2) \) in \( S_1 \). \[ (x_1, kx_1) + (x_2, kx_2) = (x_1 + x_2, kx_1 + kx_2) = (x_1 + x_2, k(x_1 + x_2)) \] This sum is also in \( S_1 \) since it's of the form \( (x, kx) \). #### **Closure under Scalar Multiplication** Take any scalar \( a \in \mathbb{R} \) and \( (x, kx) \in S_1 \): \[ a(x, kx) = (ax, a(kx)) = (ax, k(ax)) \] This result is also in \( S_1 \). **Explanation:** Both closure properties (addition and scalar multiplication) hold, and the zero vector \((0, 0)\) is in \( S_1 \). All other vector space axioms (associativity, commutativity, distributivity, existence of additive identity and inverses, etc.) are satisfied because \( S_1 \) is a subspace of \( \mathbb{R}^2 \). --- ### **Final Answer** **Yes, \( S_1 \) is a vector space under regular vector addition and scalar multiplication, because it satisfies all ten axioms of a vector space.**