Straight Lines
Geometric Loci
Grade 11
Question:
<p>If one diagonal of a square is the portion of the line <span class="math">\frac{x}{a} + \frac{y}{b} = 1</span> intercepted by the axes, then the extremities of the other diagonal of the square are:</p>
<p>(a) <span class="math">\left(\frac{a+b}{2}, \frac{a+b}{2}\right)</span></p>
<p>(b) <span class="math">\left(\frac{a-b}{2}, \frac{a+b}{2}\right)</span></p>
<p>(c) <span class="math">\left(\frac{a-b}{2}, \frac{b-a}{2}\right)</span></p>
<p>(d) <span class="math">\left(\frac{a+b}{2}, \frac{b-a}{2}\right)</span></p>
Step-by-Step Solution
Key Concept: The diagonals of a square bisect each other at right angles; use the perpendicularity condition to find the other diagonal endpoints.
<p><strong>Step 1:</strong> The line <span class="math">\frac{x}{a} + \frac{y}{b} = 1</span> intersects the axes at A(a, 0) and B(0, b).</p><p><strong>Step 2:</strong> One diagonal of the square is AB with endpoints (a, 0) and (0, b).</p><p><strong>Step 3:</strong> The center of the square is the midpoint of AB: <span class="math">M = \left(\frac{a}{2}, \frac{b}{2}\right)</span></p><p><strong>Step 4:</strong> For a square, the diagonals are perpendicular and bisect each other at M.</p><p><strong>Step 5:</strong> The other diagonal is perpendicular to AB. The direction vector of AB is (-a, b).</p><p><strong>Step 6:</strong> A perpendicular direction is (b, a). The other diagonal endpoints are at distance <span class="math">\frac{\sqrt{a^2+b^2}}{2}</span> from M along this direction.</p><p><strong>Step 7:</strong> The extremities are <span class="math">\left(\frac{a-b}{2}, \frac{b-a}{2}\right)</span> and <span class="math">\left(\frac{a+b}{2}, \frac{b+a}{2}\right)</span></p><p>∴ Answer is (c).</p>
Correct Answer: C