Straight Lines
Locus of centroid
Grade 11
Question:
<p>If the vertices <span class="math inline">\(P\)</span> and <span class="math inline">\(Q\)</span> of a triangle <span class="math inline">\(PQR\)</span> are given by <span class="math inline">\((2, 5)\)</span> and <span class="math inline">\((4, -11)\)</span> respectively, and the point <span class="math inline">\(R\)</span> moves along the line <span class="math inline">\(N: 9x + 7y + 4 = 0\)</span>, then the locus of the centroid of the triangle <span class="math inline">\(PQR\)</span> is a straight line parallel to:</p>
<p>(a) PQ</p>
<p>(b) QR</p>
<p>(c) RP</p>
<p>(d) N</p>
Step-by-Step Solution
Key Concept: The locus of the centroid is obtained by expressing the centroid coordinates in terms of the moving point R, then eliminating R to get the locus equation.
<p><strong>Solution:</strong> The centroid of triangle <span class="math inline">\(PQR\)</span> is <span class="math inline">\(G = \left(\frac{2 + 4 + x_R}{3}, \frac{5 + (-11) + y_R}{3}\right) = \left(\frac{6 + x_R}{3}, \frac{-6 + y_R}{3}\right)\)</span>. Since <span class="math inline">\(R(x_R, y_R)\)</span> moves on the line <span class="math inline">\(9x_R + 7y_R + 4 = 0\)</span>, we have <span class="math inline">\(y_R = -\frac{9x_R + 4}{7}\)</span>. Let <span class="math inline">\(G = (h, k)\)</span>. Then <span class="math inline">\(x_R = 3h - 6\)</span> and <span class="math inline">\(y_R = 3k + 6\)</span>. Substituting into the line equation: <span class="math inline">\(9(3h - 6) + 7(3k + 6) + 4 = 0\)</span>, which gives <span class="math inline">\(27h + 21k - 54 + 42 + 4 = 0\)</span>, so <span class="math inline">\(27h + 21k - 8 = 0\)</span> or <span class="math inline">\(9h + 7k - \frac{8}{3} = 0\)</span>. This is parallel to <span class="math inline">\(9x + 7y + 4 = 0\)</span> (same coefficients for x and y).</p>
Correct Answer: D