Straight Lines
Orthocentre and triangle
Grade 11
Question:
<p>Let the equations of two sides of a triangle be \(3x - 2y + 6 = 0\) and \(4x + 5y - 20 = 0\). If the orthocentre of this triangle is at (1, 1), then the equation of its third side is:</p>
<p>\(122y - 26x - 1675 = 0\)</p>
<p>\(122y + 26x + 1675 = 0\)</p>
<p>\(26x + 61y + 1675 = 0\)</p>
<p>\(26x - 122y - 1675 = 0\)</p>
Step-by-Step Solution
Key Concept: The third side must be perpendicular to the altitude from the opposite vertex. Since the orthocentre is the intersection of altitudes, the altitude from the vertex on the third side is perpendicular to that side and passes through (1,1).
<p><strong>Step 1:</strong> Find the slopes of given sides.</p><p>Line 1: 3x - 2y + 6 = 0 → slope m₁ = 3/2</p><p>Line 2: 4x + 5y - 20 = 0 → slope m₂ = -4/5</p><p><strong>Step 2:</strong> The altitude to Line 1 (perpendicular to it) has slope = -2/3 and passes through H(1,1):</p><p>y - 1 = -2/3(x - 1) → 2x + 3y - 5 = 0</p><p><strong>Step 3:</strong> The altitude to Line 2 (perpendicular to it) has slope = 5/4 and passes through H(1,1):</p><p>y - 1 = 5/4(x - 1) → 5x - 4y - 1 = 0</p><p><strong>Step 4:</strong> The third side is perpendicular to the altitude from its opposite vertex. Since altitudes from the third side's endpoints lie along 2x + 3y - 5 = 0 and 5x - 4y - 1 = 0, the third side must be perpendicular to the line joining these altitude directions.</p><p><strong>Step 5:</strong> Find vertex A (intersection of Line 1 and altitude to Line 2): 3x - 2y + 6 = 0 and 5x - 4y - 1 = 0 → A(-8, -9)</p><p><strong>Step 6:</strong> Find vertex B (intersection of Line 2 and altitude to Line 1): 4x + 5y - 20 = 0 and 2x + 3y - 5 = 0 → B(5, 0)</p><p><strong>Step 7:</strong> The third side passes through A(-8, -9) and B(5, 0):</p><p>Slope = (0-(-9))/(5-(-8)) = 9/13</p><p>Equation: y - 0 = 9/13(x - 5) → 9x - 13y - 45 = 0</p><p>∴ Answer: 9x - 13y - 45 = 0</p>
Correct Answer: D