## Precalculus (6th Edition) Blitzer

Step 1. Using the distance formula, we have $\bar AB=\sqrt {(3-1)^2+(3+d-1-d)^2}=\sqrt {4+4}=2\sqrt 2$ Step 2. Similarly, we have $\bar BC=\sqrt {(6-3)^2+(6+d-3-d)^2}=\sqrt {9+9}=3\sqrt 2$ Step 3. We have $\bar AC=\sqrt {(6-1)^2+(6+d-1-d)^2}=\sqrt {25+25}=5\sqrt 2$ Step 4. Since $\bar AC=\bar AB +\bar BC$, we conclude that points A, B, C are collinear.