Problem 72 - College Algebra

Question

Show that the points \(A(1,1+d), B(3,3+d),\) and \(C(6,6+d)\) are collinear (lie along a straight line) by showing that the distance from \(A\) to \(B\) plus the distance from \(B\) to \(C\) equals the distance from \(A\) to \(C\).

Step-by-Step Solution

Answer

The points A, B, and C are collinear, as the sum of distances between A and B and B and C equals the distance between A and C.

Step 1: Compute Distance between A and B

First, calculate the distance between points A and B. The distance 'd' between any two points (x1,y1) and (x2,y2) is given by the formula \(d = \sqrt{(x2-x1)^{2} + (y2-y1)^{2}}\). Accordingly, the distance AB between points A(1,1+d) and B(3,3+d) would be \(AB = \sqrt{(3-1)^2+(3+d-(1+d))^2} = \sqrt{4 + 4} = 2\sqrt{2}\)

Step 2: Compute Distance between B and C

Next, calculate the distance BC between points B(3,3+d) and C(6,6+d). Similar to step 1, we apply the distance formula to obtain: \(BC = \sqrt{(6-3)^2 + (6+d-(3+d))^2} = 3\sqrt{2}\)

Step 3: Compute Distance between A and C

Lastly, calculate the distance AC between points A(1,1+d) and C(6,6+d) using the distance formula. Hence, \(AC = \sqrt{(6-1)^2+(6+d-(1+d))^2} = 5\sqrt{2}\)

Step 4: Verify Collinearity

Now, the final step is to prove the points are collinear. The points on a straight line are collinear if the sum of distance of two pairs equals the distance between the farthest pair. Here, we add the distances AB and BC from Step 1 and Step 2: \(AB + BC = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}\). Notice that this is indeed equal to the distance AC from Step 3. Consequently, it validates that points A, B, and C are collinear.