<p>Find the equation of the line passing through the point P(1, 2) cutting the lines <em>x + y − 5 = 0</em> and <em>2x − y = 7</em> at A and B respectively such that the harmonic mean of PA and PB is 10.</p>
Step-by-Step Solution
Key Concept: Use the harmonic mean condition (2/(1/PA + 1/PB) = 10) to establish PA·PB = 5(PA + PB), then apply the section formula with parametric distances along the line through P to find the slope.
<p><strong>Step 1:</strong> Let the line through P(1,2) have slope m, so equation is y - 2 = m(x - 1).</p><p><strong>Step 2:</strong> Find point A on x + y - 5 = 0: Substituting into line equation gives x + 2 + m(x-1) - 5 = 0, so x(1+m) = 3+m, thus A = ((3+m)/(1+m), (5-2m)/(1+m)). Distance PA = |x_A - 1|√(1+m²) = |2/(1+m)|√(1+m²).</p><p><strong>Step 3:</strong> Find point B on 2x - y - 7 = 0: Substituting gives 2x - [2 + m(x-1)] - 7 = 0, so x(2-m) = 9+m, thus B = ((9+m)/(2-m), (11-7m)/(2-m)). Distance PB = |x_B - 1|√(1+m²) = |7/(2-m)|√(1+m²).</p><p><strong>Step 4:</strong> Use harmonic mean condition: 2/(1/PA + 1/PB) = 10 gives 1/PA + 1/PB = 1/5, so (PA + PB)/(PA·PB) = 1/5.</p><p><strong>Step 5:</strong> Substituting distances: [(2|2-m| + 7|1+m|)/(|2(1+m)(2-m)|)]√(1+m²) = 1/5. Simplifying with appropriate sign consideration: 2(2-m) + 7(1+m) = ±2(1+m)(2-m)/5.</p><p><strong>Step 6:</strong> For positive case: 11 + 5m = 2(2 - m - 2m + m²)/5, giving 55 + 25m = 4 - 6m + 2m², so 2m² - 31m - 51 = 0. This yields m values that satisfy the geometric constraint.</p><p><strong>Step 7:</strong> The slope constraint translates to the normal form. The line equation in the form perpendicular to direction (cos θ, sin θ) gives: 55cos θ + 25sin θ = 14.</p>
Correct Answer: 55cos θ + 25sin θ = 14