Straight Lines
Distance from a line
Grade 11
Question:
<p>Let \( A(x_1, y_1) \), \( B(x_2, y_2) \) and \( C(x_3, y_3) \) be the vertices of a triangle such that algebraic sum of perpendicular distance from \( A \), \( B \) and \( C \) to the variable line \( ax + by + c = 0 \) is always '0'. If \( 3a + 2b + c = 0 \), then find the value of \( \displaystyle\sum_{i=1}^{3} (x_i + y_i) \).</p>
Step-by-Step Solution
Key Concept: If the algebraic sum of perpendicular distances from three fixed points to a variable line is always zero, the line must pass through the centroid of the triangle. Use this geometric constraint along with the given condition to find the sum of coordinates.
<p><strong>Step 1: Set up the algebraic sum of distances</strong></p><p>The signed perpendicular distance from point (x_i, y_i) to line ax + by + c = 0 is: d_i = (ax_i + by_i + c)/√(a² + b²)</p><p>Given: d₁ + d₂ + d₃ = 0 (for all values of a, b, c satisfying the constraint)</p><p><strong>Step 2: Express the constraint</strong></p><p>Algebraic sum = [a(x₁ + x₂ + x₃) + b(y₁ + y₂ + y₃) + 3c]/√(a² + b²) = 0</p><p>Since √(a² + b²) ≠ 0:</p><p>a(x₁ + x₂ + x₃) + b(y₁ + y₂ + y₃) + 3c = 0</p><p><strong>Step 3: Analyze the constraint equation</strong></p><p>This must hold for ALL values of a, b, c satisfying 3a + 2b + c = 0.</p><p>From 3a + 2b + c = 0, we get: c = -3a - 2b</p><p>Substituting:</p><p>a(x₁ + x₂ + x₃) + b(y₁ + y₂ + y₃) + 3(-3a - 2b) = 0</p><p>a(x₁ + x₂ + x₃ - 9) + b(y₁ + y₂ + y₃ - 6) = 0</p><p><strong>Step 4: Apply independence condition</strong></p><p>Since this holds for all independent values of a and b, both coefficients must be zero:</p><p>x₁ + x₂ + x₃ = 9</p><p>y₁ + y₂ + y₃ = 6</p><p><strong>Step 5: Calculate the sum</strong></p><p>∑(x_i + y_i) = (x₁ + x₂ + x₃) + (y₁ + y₂ + y₃) = 9 + 6 = 15</p><p><strong>∴ Answer: 15</strong></p>
Correct Answer: 15