Straight Lines
Area of Triangle and Orthocentre
Grade 11
Question:
<p>Let <i>k</i> be an integer such that the triangle with vertices <i>(k, −3k)</i>, <i>(5, k)</i> and <i>(−k, 2)</i> has area 28 sq. units. Then the orthocentre of this triangle is at the point:</p>
<p>\(\left(1, \dfrac{3}{4}\right)\)</p>
<p>\(\left(1, -\dfrac{3}{4}\right)\)</p>
<p>\(\left(2, \dfrac{1}{2}\right)\)</p>
<p>\(\left(2, -\dfrac{1}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: Use the area formula with determinant to find k, then calculate the orthocenter by finding the intersection of altitudes (using perpendicularity condition: product of slopes = -1).
<p><strong>Step 1: Find k using area formula</strong></p><p>Area = ½|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|</p><p>28 = ½|k(k - 2) + 5(2 + 3k) + (-k)(-3k - k)|</p><p>56 = |k² - 2k + 10 + 15k + 4k²|</p><p>56 = |5k² + 13k + 10|</p><p>This gives: 5k² + 13k + 10 = ±56</p><p><strong>Case 1:</strong> 5k² + 13k - 46 = 0 → k = 2 or k = -4.6 (reject non-integer)</p><p><strong>Case 2:</strong> 5k² + 13k + 66 = 0 → No real solutions</p><p>∴ k = 2</p><p><strong>Step 2: Find vertices with k = 2</strong></p><p>A(2, -6), B(5, 2), C(-2, 2)</p><p><strong>Step 3: Find orthocenter</strong></p><p>Slope of BC: (2-2)/(5-(-2)) = 0 (horizontal line)</p><p>Altitude from A is perpendicular to BC: vertical line x = 2</p><p>Slope of AB: (2-(-6))/(5-2) = 8/3</p><p>Altitude from C perpendicular to AB: slope = -3/8</p><p>Line through C(-2, 2): y - 2 = -3/8(x + 2)</p><p>At x = 2: y - 2 = -3/8(4) = -3/2 → y = 1/2</p><p>∴ Orthocenter: (2, 1/2)</p>
Correct Answer: A