Straight Lines
Perpendicular Lines and Intersection
Grade 11
Question:
<p>Given that lines \(x + (a-1)y = 1\) and \(2x + a^2 y = 1\) are perpendicular, then the product of slopes of lines is \(-1\). Find the distance from the origin to the point of intersection of the lines (taking the real value of \(a\)).</p>
Step-by-Step Solution
Key Concept: Two lines are perpendicular when the product of their slopes equals -1. Use this condition to find the value of parameter 'a', then find the intersection point and calculate its distance from the origin.
<p><strong>Step 1: Find slopes from standard form</strong></p><p>Line 1: x + (a-1)y = 1 → slope m₁ = -1/(a-1)</p><p>Line 2: 2x + a²y = 1 → slope m₂ = -2/a²</p><p><strong>Step 2: Apply perpendicularity condition m₁·m₂ = -1</strong></p><p>[-1/(a-1)]·[-2/a²] = -1</p><p>2/[(a-1)a²] = -1</p><p>2 = -(a-1)a²</p><p>2 = -a³ + a²</p><p>a³ - a² + 2 = 0</p><p><strong>Step 3: Solve cubic equation</strong></p><p>Testing values: a = -1 works</p><p>(-1)³ - (-1)² + 2 = -1 - 1 + 2 = 0 ✓</p><p>Factoring: (a+1)(a² - 2a + 2) = 0</p><p>From a² - 2a + 2 = 0: Δ = 4 - 8 = -4 < 0 (complex roots)</p><p>∴ Real value: a = -1</p><p><strong>Step 4: Find intersection point with a = -1</strong></p><p>Line 1: x - 2y = 1</p><p>Line 2: 2x + y = 1</p><p>From Line 1: x = 1 + 2y</p><p>Substituting: 2(1 + 2y) + y = 1 → 2 + 4y + y = 1 → 5y = -1 → y = -1/5</p><p>x = 1 + 2(-1/5) = 1 - 2/5 = 3/5</p><p>Intersection: (3/5, -1/5)</p><p><strong>Step 5: Find distance from origin</strong></p><p>d = √[(3/5)² + (-1/5)²] = √[9/25 + 1/25] = √(10/25) = √10/5 = √10/5 ≈ 0.6324</p><p>∴ Answer: 0.6324</p>
Correct Answer: 0.6324