1. The problem states that point Q is the midpoint of line PR, and we are given points P(-4, 6) and Q(k, -2). We need to find the value of $k$ given that the distance $PQ = 17$ units. 2. The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. Applying the distance formula to points $P(-4, 6)$ and $Q(k, -2)$: $$17 = \sqrt{(k - (-4))^2 + (-2 - 6)^2} = \sqrt{(k + 4)^2 + (-8)^2} = \sqrt{(k + 4)^2 + 64}$$ 4. Square both sides to eliminate the square root: $$17^2 = (k + 4)^2 + 64$$ $$289 = (k + 4)^2 + 64$$ 5. Subtract 64 from both sides: $$289 - 64 = (k + 4)^2$$ $$225 = (k + 4)^2$$ 6. Take the square root of both sides: $$k + 4 = \pm 15$$ 7. Solve for $k$: - If $k + 4 = 15$, then $k = 11$ - If $k + 4 = -15$, then $k = -19$ 8. Since the problem context suggests $k$ is positive and matches the options, the value of $k$ is $11$. **Final answer:** $k = 11$