Question

If the algebraic sum of the distances from the points $(2, 0)$, $(0, 2)$, and $(1, 1)$ to a variable straight line is zero, then the line passes through the fixed point.

• A. (-1, 1)
• B. ( 1, -1)
• C. ( -1, -1)
• D. ( 1, 1)

Answer 1

The Correct Option is D

Given: The algebraic sum of the distances from the points \( (2, 0) \), \( (0, 2) \), and \( (1, 1) \) to a variable straight line is zero. We are tasked with finding the fixed point through which this line passes. Step 1: Understanding the condition. The condition that the algebraic sum of the distances from a set of points to a straight line is zero implies that the points are symmetrically distributed with respect to the line. The algebraic sum means that the signed distances from the points to the line are being summed, which suggests that the line divides the region formed by the points such that the total positive and negative distances balance out to zero. Step 2: General form of the line. The equation of a straight line in the plane can be written as: $$ ax + by + c = 0 $$ The distance \( d \) from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) is given by the formula: $$ d = \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} $$ However, we are interested in the condition that the sum of the distances from the points \( (2, 0) \), \( (0, 2) \), and \( (1, 1) \) to the line is zero. Step 3: Interpret the geometric meaning. The fact that the algebraic sum of the distances is zero suggests that the line passes through the centroid of the points. The centroid \( G \) of the points \( (2, 0) \), \( (0, 2) \), and \( (1, 1) \) is the average of their coordinates: $$ G = \left( \frac{2 + 0 + 1}{3}, \frac{0 + 2 + 1}{3} \right) = \left( \frac{3}{3}, \frac{3}{3} \right) = (1, 1) $$ Therefore, the line must pass through the point \( (1, 1) \). Conclusion: The fixed point through which the line passes is \( \boxed{(1, 1)} \).