Straight Lines
Division of quadrilaterals
Grade 11
Question:
<p>Vertices of a parallelogram ABCD are <span class="math inline">\(A(3, 1)\)</span>, <span class="math inline">\(B(13, 6)\)</span>, <span class="math inline">\(C(13, 21)\)</span> and <span class="math inline">\(D(3, 16)\)</span>. If a line passing through the origin divides the parallelogram into two congruent parts then the slope of the line is:</p>
<p>(a) <span class="math inline">\(\frac{11}{12}\)</span></p>
<p>(b) <span class="math inline">\(\frac{11}{8}\)</span></p>
<p>(c) <span class="math inline">\(\frac{25}{8}\)</span></p>
<p>(d) <span class="math inline">\(\frac{13}{8}\)</span></p>
Step-by-Step Solution
Key Concept: A line through the origin divides a parallelogram into two congruent parts if and only if it passes through the center of the parallelogram. The center is the intersection point of the diagonals.
<p><strong>Step 1: Find the center of the parallelogram.</strong></p><p>In a parallelogram, the diagonals bisect each other. The center is the midpoint of diagonal AC (or equivalently, diagonal BD).</p><p>Midpoint of AC: $\left(\frac{3+13}{2}, \frac{1+21}{2}\right) = \left(\frac{16}{2}, \frac{22}{2}\right) = (8, 11)$</p><p><strong>Step 2: Verify using diagonal BD.</strong></p><p>Midpoint of BD: $\left(\frac{13+3}{2}, \frac{6+16}{2}\right) = \left(\frac{16}{2}, \frac{22}{2}\right) = (8, 11)$ ✓</p><p><strong>Step 3: Find the slope of the line through origin and center.</strong></p><p>The line passes through origin $O(0, 0)$ and center $P(8, 11)$.</p><p>Slope $= \frac{11-0}{8-0} = \frac{11}{8}$</p><p><strong>Step 4: Verify this divides the parallelogram into congruent parts.</strong></p><p>Since the line passes through the center of the parallelogram, it creates two halves that are symmetric about this center point, making them congruent by point symmetry.</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B