<p>If the vertices P and Q of a triangle PQR are given by <span class="math inline">\((2, 5)\)</span> and <span class="math inline">\((4, -11)\)</span> respectively, and the point R moves along the line N: <span class="math inline">\(9x + 7y + 4 = 0\)</span>, then the locus of the centroid of the triangle PQR is a straight line parallel to:</p>
Step-by-Step Solution
Key Concept: The centroid of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3). Since R moves on a line, the locus of the centroid will also be a line parallel to that line.
<p><strong>Step 1:</strong> Identify the fixed vertices and the moving vertex.</p><p>P = (2, 5), Q = (4, -11), and R = (x, y) moves on line N: 9x + 7y + 4 = 0</p><p><strong>Step 2:</strong> Write the formula for the centroid G of triangle PQR.</p><p>G = ((2 + 4 + x)/3, (5 - 11 + y)/3) = ((6 + x)/3, (-6 + y)/3)</p><p><strong>Step 3:</strong> Let the centroid G = (h, k).</p><p>h = (6 + x)/3 ⟹ x = 3h - 6</p><p>k = (-6 + y)/3 ⟹ y = 3k + 6</p><p><strong>Step 4:</strong> Since R(x, y) lies on line N, substitute the parametric relations into the equation of N.</p><p>9x + 7y + 4 = 0</p><p>9(3h - 6) + 7(3k + 6) + 4 = 0</p><p>27h - 54 + 21k + 42 + 4 = 0</p><p>27h + 21k - 8 = 0</p><p><strong>Step 5:</strong> Replace (h, k) with (x, y) to get the locus equation.</p><p>27x + 21y - 8 = 0</p><p>Dividing by 3: 9x + 7y - 8/3 = 0</p><p><strong>Step 6:</strong> Compare with the original line N: 9x + 7y + 4 = 0.</p><p>Both equations have the same coefficients for x and y (9 and 7), meaning the lines have the same slope. Therefore, the locus of the centroid is parallel to line N.</p><p><strong>∴ Answer:</strong> D</p>
Correct Answer: D