Straight Lines
Distance from a Point to a Line
Grade 11
Question:
<p><strong>251.</strong> Let \(f(x, y)\) be a locus of a point \(P(x, y)\) satisfying \(\alpha(2x - y + 1) + \beta(3x - y) + \gamma(2x + y - 5) = 0\) \(\forall\, \alpha, \beta, \gamma \in R\). The least distance between the curve \(f(x, y)\) and straight line \(3x - 4y + 19 = 0\) is:</p>
<p>(a) 3</p>
<p>(b) 2</p>
<p>(c) \(\dfrac{7}{5}\)</p>
<p>(d) \(\dfrac{26}{5}\)</p>
Step-by-Step Solution
Key Concept: For the equation to hold for all values of α, β, γ ∈ ℝ, each coefficient must independently equal zero. This means finding the intersection point of the three lines represented by the three linear expressions.
<p><strong>Step 1:</strong> Since α(2x - y + 1) + β(3x - y) + γ(2x + y - 5) = 0 for all α, β, γ ∈ ℝ, each coefficient must equal zero:</p><p>• 2x - y + 1 = 0 ... (1)</p><p>• 3x - y = 0 ... (2)</p><p>• 2x + y - 5 = 0 ... (3)</p><p><strong>Step 2:</strong> From equation (2): y = 3x</p><p><strong>Step 3:</strong> Substitute into equation (1): 2x - 3x + 1 = 0 → x = 1, y = 3</p><p><strong>Step 4:</strong> Verify in equation (3): 2(1) + 3 - 5 = 0 ✓</p><p>So f(x, y) represents the point P(1, 3)</p><p><strong>Step 5:</strong> Find distance from P(1, 3) to line 3x - 4y + 19 = 0:</p><p>Distance = |3(1) - 4(3) + 19|/√(9 + 16) = |3 - 12 + 19|/5 = |10|/5 = 2</p><p><strong>∴ Answer: B (least distance = 2)</strong></p>
Correct Answer: B