Straight Lines
Distance from a point to a line
Grade 11
Question:
<p>Let the position of a bungalow be \(P(x_1, y_1)\). Given \(PM = 100\) and \(PN = 100\) where \(M\) and \(N\) are feet of perpendiculars from \(P\) to the lines \(x + y = 8\) and \(y - x = 6\) respectively. Then: \[\frac{x_1 + y_1 - 8}{\sqrt{2}} = \pm 100 \quad \text{and} \quad \frac{x_1 - y_1 + 6}{\sqrt{2}} = \pm 100\] Which of the following are possible locations of the bungalow?</p>
<p>\(\left(1 + 100\sqrt{2},\ 7\right)\)</p>
<p>\(\left(1 - 100\sqrt{2},\ 7\right)\)</p>
<p>\(\left(1,\ 7 + 100\sqrt{2}\right)\)</p>
<p>\(\left(1,\ 7 - 100\sqrt{2}\right)\)</p>
Step-by-Step Solution
Key Concept: The perpendicular distance from a point P(x₁, y₁) to a line ax + by + c = 0 is |ax₁ + by₁ + c|/√(a² + b²). We use this formula with the constraint that PM = PN = 100 to establish two equations with ± signs, then verify which given points satisfy both equations.
<p><strong>Step 1: Set up the distance equations</strong></p><p>For line x + y − 8 = 0, the perpendicular distance is:</p><p>PM = |x₁ + y₁ − 8|/√2 = 100</p><p>This gives: x₁ + y₁ − 8 = ±100√2</p><p><strong>Step 2: Set up the second distance equation</strong></p><p>For line −x + y − 6 = 0 (or y − x − 6 = 0), the perpendicular distance is:</p><p>PN = |−x₁ + y₁ − 6|/√2 = 100</p><p>This gives: −x₁ + y₁ − 6 = ±100√2, or equivalently x₁ − y₁ + 6 = ±100√2</p><p><strong>Step 3: Identify the four possible cases</strong></p><p>From the two equations, we have four combinations of signs:</p><p><strong>Case 1:</strong> x₁ + y₁ − 8 = +100√2 and x₁ − y₁ + 6 = +100√2</p><p>Adding: 2x₁ − 2 = 200√2 ⟹ x₁ = 1 + 100√2</p><p>Subtracting: 2y₁ − 14 = 0 ⟹ y₁ = 7</p><p>Solution: (1 + 100√2, 7) ✓ <strong>Option A</strong></p><p><strong>Case 2:</strong> x₁ + y₁ − 8 = −100√2 and x₁ − y₁ + 6 = −100√2</p><p>Adding: 2x₁ − 2 = −200√2 ⟹ x₁ = 1 − 100√2</p><p>Subtracting: 2y₁ − 14 = 0 ⟹ y₁ = 7</p><p>Solution: (1 − 100√2, 7) ✓ <strong>Option B</strong></p><p><strong>Case 3:</strong> x₁ + y₁ − 8 = +100√2 and x₁ − y₁ + 6 = −100√2</p><p>Adding: 2x₁ − 2 = 0 ⟹ x₁ = 1</p><p>Subtracting: 2y₁ − 14 = 200√2 ⟹ y₁ = 7 + 100√2</p><p>Solution: (1, 7 + 100√2) ✓ <strong>Option C</strong></p><p><strong>Case 4:</strong> x₁ + y₁ − 8 = −100√2 and x₁ − y₁ + 6 = +100√2</p><p>Adding: 2x₁ − 2 = 0 ⟹ x₁ = 1</p><p>Subtracting: 2y₁ − 14 = −200√2 ⟹ y₁ = 7 − 100√2</p><p>Solution: (1, 7 − 100√2) ✓ <strong>Option D</strong></p><p><strong>Step 4: Verify all options</strong></p><p>All four combinations of signs yield valid solutions that satisfy both distance equations with PM = PN = 100.</p><p>∴ Answer: A, B, C, D</p>
Correct Answer: A, B, C, D