<p>Two vertices of a triangle are \((0, 2)\) and \((4, 3)\) and its orthocenter is at the origin. Then the third vertex \(A(a, b)\) lies in which quadrant?</p>
Step-by-Step Solution
Key Concept: The orthocenter is where altitudes meet. If the orthocenter is at the origin, then the altitude from the third vertex must pass through the origin, and the line joining the two given vertices must be perpendicular to the line from the origin to the third vertex.
<p><strong>Step 1:</strong> Let the two given vertices be B(0, 2) and C(4, 3), and the third vertex be A(a, b). The orthocenter H is at origin O(0, 0).</p><p><strong>Step 2:</strong> The altitude from A is perpendicular to BC. Since this altitude passes through both A and O(orthocenter), the line AO must be perpendicular to BC.</p><p>Slope of BC = (3-2)/(4-0) = 1/4</p><p>Slope of AO = b/a</p><p>For perpendicularity: (b/a) × (1/4) = -1 → <strong>b = -4a</strong> ... (1)</p><p><strong>Step 3:</strong> The altitude from B is perpendicular to AC. Since this altitude passes through B(0, 2) and O(0, 0), the altitude BO is the y-axis (vertical line).</p><p>For BO ⊥ AC, the line AC must be horizontal, so slope of AC = 0.</p><p>Slope of AC = (b - 3)/(a - 4) = 0 → <strong>b = 3</strong> ... (2)</p><p><strong>Step 4:</strong> From (1) and (2): 3 = -4a → a = -3/4</p><p>So A(-3/4, 3) lies in <strong>Quadrant II</strong> (negative x, positive y).</p><p>∴ Answer: B</p>
Correct Answer: B