Straight Lines
Distance between point and 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, the point P(x,y) must lie on the intersection of all three lines simultaneously. This occurs only at the point where all three lines are concurrent.
<p><strong>Step 1: Find the point of concurrency</strong></p><p>For the equation α(2x - y + 1) + β(3x - y) + γ(2x + y - 5) = 0 to hold for all α, β, γ ∈ ℝ, we need:</p><p>2x - y + 1 = 0 ... (1)<br>3x - y = 0 ... (2)<br>2x + y - 5 = 0 ... (3)</p><p><strong>Step 2: Solve the system</strong></p><p>From (2): y = 3x</p><p>Substitute in (1): 2x - 3x + 1 = 0 ⟹ x = 1</p><p>Therefore: y = 3</p><p>Verify in (3): 2(1) + 3 - 5 = 0 ✓</p><p><strong>Step 3: Calculate distance from point (1, 3) to line 3x - 4y + 19 = 0</strong></p><p>Distance = |3(1) - 4(3) + 19|/√(3² + 4²)</p><p>= |3 - 12 + 19|/√25</p><p>= |10|/5</p><p>= 2</p><p>∴ Answer: B</p>
Correct Answer: B