Let's solve the problem in **2 steps** with explanations and a final answer: --- ### **Step 1: Understand the Set and Select a Vector** **Set Definition:** - The solution set \( S_2 \) of \( y = kx + n_1 \) is \[ S_2 = \{(x, y) \in \mathbb{R}^2 \mid y = kx + n_1\} \] **Select a vector from \( S_2 \):** - Let's pick \( (x_1, y_1) \) where \( y_1 = kx_1 + n_1 \). - For example, let \( x_1 = 0 \). Then \( y_1 = n_1 \). - So, \( (0, n_1) \) is in \( S_2 \). **Explanation:** - We have chosen a vector \( (0, n_1) \) from the set \( S_2 \), which satisfies the equation \( y = kx + n_1 \). --- ### **Step 2: Check if the Zero Vector is in \( S_2 \) (Axiom)** **Zero vector in \( \mathbb{R}^2 \) is \( (0, 0) \).** - For \( (0, 0) \) to be in \( S_2 \), it must satisfy \( y = kx + n_1 \). - Substitute \( (0, 0) \): \( 0 = k \cdot 0 + n_1 \implies 0 = n_1 \) **If \( n_1 \neq 0 \), then \( (0, 0) \notin S_2 \).** **Explanation:** - The zero vector is a required element for a vector space. If \( n_1 \neq 0 \), the zero vector is not in \( S_2 \). Thus, at least one vector space axiom fails. --- ### **Final Answer:** **The set \( S_2 \) is not a vector space (for \( n_1 \neq 0 \)) because it does not contain the zero vector, violating a fundamental axiom of vector spaces.**