Straight Lines
Distance and Slope
Grade 11
Question:
<p>Given line <em>x</em> + <em>y</em> = 7 and point P(2, 3). Let point on line <em>x</em> + <em>y</em> = 7 where we draw perpendicular to point P(2, 3) be B(<em>x</em><sub>1</sub>, <em>y</em><sub>1</sub>) and point on line <em>x</em> + <em>y</em> = 7 from where point P(2, 3) is at distance 4 units be A(<em>x</em><sub>2</sub>, <em>y</em><sub>2</sub>). Find the slope of the line PA.</p>
Step-by-Step Solution
Key Concept: Find point A on line x+y=7 at distance 4 from P(2,3) using distance formula, then calculate slope. The distance formula gives a quadratic equation yielding two possible points; select the appropriate one based on context.
<p><strong>Step 1:</strong> Point A lies on x+y=7, so A(x₂, y₂) satisfies y₂ = 7-x₂</p><p><strong>Step 2:</strong> Distance PA = 4, so: (x₂-2)² + (7-x₂-3)² = 16<br>(x₂-2)² + (4-x₂)² = 16<br>x₂² - 4x₂ + 4 + x₂² - 8x₂ + 16 = 16<br>2x₂² - 12x₂ + 4 = 0<br>x₂² - 6x₂ + 2 = 0</p><p><strong>Step 3:</strong> Using quadratic formula: x₂ = (6 ± √(36-8))/2 = (6 ± √28)/2 = 3 ± √7<br>So x₂ = 3 + √7 ≈ 5.646 or x₂ = 3 - √7 ≈ 0.354</p><p><strong>Step 4:</strong> For x₂ = 3 - √7: y₂ = 7 - (3-√7) = 4 + √7<br>Slope = (4+√7-3)/(3-√7-2) = (1+√7)/(1-√7)<br>Rationalizing: [(1+√7)(1+√7)]/[(1-√7)(1+√7)] = (1+2√7+7)/(1-7) = (8+2√7)/(-6) = -(4+√7)/3<br>= -(4+2.646)/3 ≈ -6.646/3 ≈ -2.215</p><p><strong>Step 5:</strong> For x₂ = 3 + √7: y₂ = 4 - √7<br>Slope = (4-√7-3)/(3+√7-2) = (1-√7)/(1+√7)<br>Rationalizing: [(1-√7)(1-√7)]/[(1+√7)(1-√7)] = (1-2√7+7)/(1-7) = (8-2√7)/(-6) = -(4-√7)/3<br>= -(4-2.646)/3 ≈ -1.354/3 ≈ -0.4514</p><p><strong>∴ Answer: -0.4514</strong> (taking the second solution)</p>
Correct Answer: -0.4514