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Problem 639

# Prove that points $$\mathrm{A}(2,3), \mathrm{B}(4,4)$$, and $$\mathrm{C}(8,6)$$ are collinear.

Expert verified
The points A(2,3), B(4,4), and C(8,6) are collinear. This is proven by calculating the slopes of lines AB, AC, and BC. The slope of AB is $$\frac{1}{2}$$, calculated using the slope formula $$\frac{y_2 - y_1}{x_2 - x_1}$$ (specifically, $$\frac{4-3}{4-2} = \frac{1}{2}$$). The same calculation gives the slope of AC as $$\frac{1}{2}$$ (from $$\frac{6-3}{8-2}$$) and the slope of BC as $$\frac{1}{2}$$ (from $$\frac{6-4}{8-4}$$). As all three slopes are equal, the points are collinear.
See the step by step solution

## Step 1: Calculate the slope of line AB

: We will use the slope formula: $$\frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$$. For points A(2,3) and B(4,4), the slope of line AB is: $$\frac{4-3}{4-2} = \frac{1}{2}$$

## Step 2: Calculate the slope of line AC

: For points A(2,3) and C(8,6), the slope of line AC is: $$\frac{6-3}{8-2} = \frac{3}{6} = \frac{1}{2}$$

## Step 3: Calculate the slope of line BC

: For points B(4,4) and C(8,6), the slope of line BC is: $$\frac{6-4}{8-4} = \frac{2}{4} = \frac{1}{2}$$

## Step 4: Compare the slopes

: We have the following slopes: - Slope of line AB = $$\frac{1}{2}$$ - Slope of line AC = $$\frac{1}{2}$$ - Slope of line BC = $$\frac{1}{2}$$ All three slopes have the same value of $$\frac{1}{2}$$. Therefore, points A, B, and C are collinear.

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