<p>The foot of the perpendicular drawn from the origin, on the line, \(3x + y = \lambda\,(\lambda \ne 0)\) is \(P\). If the line meets \(x\)-axis at \(A\) and \(y\)-axis at \(B\), then the ratio \(BP : PA\) is</p>
Step-by-Step Solution
Key Concept: The foot of perpendicular from origin to line 3x + y = λ lies on the line, and we can use the parametric form of the perpendicular (in direction of normal) combined with intercept geometry to find the ratio.
<p><strong>Step 1:</strong> Find coordinates of P (foot of perpendicular from origin).</p><p>The perpendicular from origin to line 3x + y = λ has direction vector (3, 1) (normal to the line).</p><p>Parametric form: (3t, t) must satisfy 3x + y = λ</p><p>So: 3(3t) + t = λ ⟹ 10t = λ ⟹ t = λ/10</p><p>Thus P = (3λ/10, λ/10)</p><p><strong>Step 2:</strong> Find coordinates of A and B (intercepts).</p><p>For A (x-intercept, y = 0): 3x = λ ⟹ A = (λ/3, 0)</p><p>For B (y-intercept, x = 0): y = λ ⟹ B = (0, λ)</p><p><strong>Step 3:</strong> Calculate distances BP and PA.</p><p>BP² = (3λ/10 - 0)² + (λ/10 - λ)² = (3λ/10)² + (-9λ/10)² = 9λ²/100 + 81λ²/100 = 90λ²/100</p><p>BP = (3λ√10)/10</p><p>PA² = (λ/3 - 3λ/10)² + (0 - λ/10)² = (10λ - 9λ)²/900 + λ²/100 = λ²/900 + 9λ²/900 = 10λ²/900</p><p>PA = λ√10/30</p><p><strong>Step 4:</strong> Find ratio BP : PA.</p><p>BP/PA = [(3λ√10)/10] / [λ√10/30] = (3λ√10)/10 × 30/(λ√10) = 90/10 = 9/1</p><p>∴ BP : PA = <strong>9 : 1</strong></p>
Correct Answer: C