1. The problem states that point K is the image of point J under a translation vector \(\begin{pmatrix} m \\ n \end{pmatrix}\), and the distance between J and K is 13 units. 2. Given coordinates: \(J(-5, -3)\) and \(K(0, y)\). 3. The translation vector is \(\begin{pmatrix} m \\ n \end{pmatrix} = \begin{pmatrix} 0 - (-5) \\ y - (-3) \end{pmatrix} = \begin{pmatrix} 5 \\ y + 3 \end{pmatrix}\). 4. The distance \(JK = 13\) units, so using the distance formula: $$\sqrt{(0 - (-5))^2 + (y - (-3))^2} = 13$$ which simplifies to $$\sqrt{5^2 + (y + 3)^2} = 13$$ 5. Square both sides: $$25 + (y + 3)^2 = 169$$ 6. Subtract 25 from both sides: $$(y + 3)^2 = 144$$ 7. Take the square root: $$y + 3 = \pm 12$$ 8. Solve for \(y\): - If \(y + 3 = 12\), then \(y = 9\) - If \(y + 3 = -12\), then \(y = -15\) 9. The point K is at \((0, y)\), so possible K points are \((0, 9)\) or \((0, -15)\). 10. The translation vectors are: - For \(y = 9\): \(\begin{pmatrix} 5 \\ 9 + 3 \end{pmatrix} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}\) - For \(y = -15\): \(\begin{pmatrix} 5 \\ -15 + 3 \end{pmatrix} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}\) 11. From the options given: A \(\begin{pmatrix} -5 \\ 12 \end{pmatrix}\) B \(\begin{pmatrix} 5 \\ -12 \end{pmatrix}\) C \(\begin{pmatrix} 12 \\ 5 \end{pmatrix}\) D \(\begin{pmatrix} 5 \\ 12 \end{pmatrix}\) 12. The correct translation vectors are options B and D. Final answer: The translation vector \(\begin{pmatrix} m \\ n \end{pmatrix}\) is either \(\begin{pmatrix} 5 \\ 12 \end{pmatrix}\) or \(\begin{pmatrix} 5 \\ -12 \end{pmatrix}\).