Straight Lines
Distance and Altitudes
Grade 11
Question:
<p>The length of altitude through <span class="math">A</span> of <span class="math">\triangle ABC</span>, where <span class="math">A \equiv (-3, 0)</span>, <span class="math">B \equiv (4, -1)</span>, <span class="math">C \equiv (5, 2)</span>, is</p>
<p>(a) <span class="math">\frac{2}{\sqrt{10}}</span></p>
<p>(b) <span class="math">\frac{4}{\sqrt{10}}</span></p>
<p>(c) <span class="math">\frac{11}{\sqrt{10}}</span></p>
<p>(d) <span class="math">\frac{22}{\sqrt{10}}</span></p>
Step-by-Step Solution
Key Concept: The altitude from vertex A to side BC is the perpendicular distance from point A to the line containing BC. This distance equals the length of the altitude, calculated using the point-to-line distance formula.
<p><strong>Step 1: Find the equation of line BC.</strong></p><p>Given: B ≡ (4, -1) and C ≡ (5, 2)</p><p>Slope of BC: m = (2 - (-1))/(5 - 4) = 3/1 = 3</p><p>Using point-slope form with point B(4, -1):</p><p>y - (-1) = 3(x - 4)</p><p>y + 1 = 3x - 12</p><p>3x - y - 13 = 0</p><p><strong>Step 2: Apply the point-to-line distance formula.</strong></p><p>The perpendicular distance from point A(-3, 0) to line 3x - y - 13 = 0 is:</p><p>Distance = |ax₀ + by₀ + c|/√(a² + b²)</p><p>where (x₀, y₀) = A(-3, 0) and the line is 3x - y - 13 = 0</p><p>Distance = |3(-3) + (-1)(0) + (-13)|/√(3² + (-1)²)</p><p>Distance = |-9 + 0 - 13|/√(9 + 1)</p><p>Distance = |-22|/√10</p><p>Distance = 22/√10</p><p><strong>∴ Answer: d</strong></p>
Correct Answer: d