Straight Lines
Rotation of lines and triangle properties
Grade 11
Question:
<p>If the straight line \(3x - 4y + 7 = 0\) rotated through 90° about a point (3, 4) meets the coordinate axes at <em>A</em> and <em>B</em> and a \(\triangle AOB\) is formed (<em>O</em> is the origin), then:</p>
<p>(a) area of the triangle formed by orthocentre, circumcentre and centroid of the \(\triangle AOB\) is 1 sq. units.</p>
<p>(b) area of the triangle formed by orthocentre, circumcentre and incentre of the \(\triangle AOB\) is 1 sq. units.</p>
<p>(c) distance between orthocentre and incentre is 2 units.</p>
<p>(d) distance between orthocentre and circumcentre is 5 units.</p>
Step-by-Step Solution
Key Concept: When a line is rotated 90° about a point, the new line is perpendicular to the original. Use the perpendicularity condition (product of slopes = -1) and the fact that the rotated line passes through (3,4) to find its equation, then find intercepts.
<p><strong>Step 1:</strong> Find the slope of the original line 3x - 4y + 7 = 0.</p><p>Rewrite as: y = (3/4)x + 7/4, so m₁ = 3/4</p><p><strong>Step 2:</strong> The rotated line is perpendicular to the original line.</p><p>For perpendicular lines: m₁ · m₂ = -1</p><p>Therefore: m₂ = -4/3</p><p><strong>Step 3:</strong> The rotated line passes through (3, 4) with slope -4/3.</p><p>Using point-slope form: y - 4 = (-4/3)(x - 3)</p><p>3(y - 4) = -4(x - 3)</p><p>3y - 12 = -4x + 12</p><p>4x + 3y = 24</p><p><strong>Step 4:</strong> Find intercepts to locate points A and B.</p><p>At x-axis (y = 0): 4x = 24, so x = 6 → A(6, 0)</p><p>At y-axis (x = 0): 3y = 24, so y = 8 → B(0, 8)</p><p><strong>Step 5:</strong> Calculate area of △AOB with O(0,0), A(6,0), B(0,8).</p><p>Area = (1/2) × 6 × 8 = 24 square units</p><p>∴ Answer: A</p>
Correct Answer: A