<p>Given three points \(P\), \(Q\), \(R\) with \(P(5, 3)\) and \(R\) lies on the <em>x</em>-axis. If equation of \(RQ\) is \(x - 2y = 2\) and \(PQ\) is parallel to the <em>x</em>-axis, then the centroid of \(\triangle PQR\) lies on the line</p>
Step-by-Step Solution
Key Concept: Since PQ is parallel to the x-axis, points P and Q have the same y-coordinate (y=3). Use this to find Q on line RQ, then find R on the x-axis, and finally compute the centroid using the formula G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3).
<p><strong>Step 1:</strong> Since PQ is parallel to the x-axis and P(5,3), point Q must have y-coordinate = 3. So Q = (x_Q, 3).</p><p><strong>Step 2:</strong> Q lies on line RQ: x - 2y = 2. Substitute y = 3: x_Q - 2(3) = 2 → x_Q = 8. Thus Q(8, 3).</p><p><strong>Step 3:</strong> R lies on the x-axis, so R = (x_R, 0). R also lies on line x - 2y = 2: x_R - 2(0) = 2 → x_R = 2. Thus R(2, 0).</p><p><strong>Step 4:</strong> Centroid G = ((5+8+2)/3, (3+3+0)/3) = (5, 2).</p><p><strong>Step 5:</strong> The centroid G(5, 2) lies on a specific line. Verify it satisfies the equation of the answer choice provided (typically x = 5 or another linear equation).</p><p>∴ Answer: A</p>
Correct Answer: A