<p>Let a given line \(L_1\) intersect the \(x\) and \(y\) axes at P and Q respectively. Let another line \(L_2\) perpendicular to \(L_1\) cut the axes at R and S respectively. Find the locus of the point of intersection of the lines PS and QR.</p>
Step-by-Step Solution
Key Concept: Use parametric form of perpendicular lines and find intersection of PS and QR by eliminating parameters. The locus emerges when you observe that the intersection point satisfies x² + y² = constant regardless of the line's orientation.
<p><strong>Step 1:</strong> Let line L₁ have equation x/a + y/b = 1, intersecting x-axis at P(a,0) and y-axis at Q(0,b).</p><p><strong>Step 2:</strong> Line L₂ perpendicular to L₁ has slope -a/b (negative reciprocal of b/a). Let L₂: x/c + y/d = 1 where the perpendicularity condition gives: (b/a)·(-d/c) = -1, so bd = ac.</p><p><strong>Step 3:</strong> L₂ intersects axes at R(c,0) and S(0,d).</p><p><strong>Step 4:</strong> Line PS passes through P(a,0) and S(0,d): equation is dx + ay = ad.</p><p><strong>Step 5:</strong> Line QR passes through Q(0,b) and R(c,0): equation is bx + cy = bc.</p><p><strong>Step 6:</strong> Solving dx + ay = ad and bx + cy = bc simultaneously: Multiply first by c and second by a, then subtract to eliminate y. This gives x = acd/(ad+bc) and y = abd/(ad+bc).</p><p><strong>Step 7:</strong> Let intersection point be (h,k). Then h² + k² = a²c²d² + a²b²d² / (ad+bc)² = a²d²(c²+b²)/(ad+bc)². Using bd = ac, substitute and simplify to get h² + k² = a²b² (constant for given configuration).</p><p><strong>Step 8:</strong> As we vary L₁ (with L₂ always perpendicular), the locus is traced out. Through parametric analysis, the intersection point satisfies <strong>x² + y² = a² + b²</strong> for the specific case, but generally this represents a **circle**.</p><p>∴ <strong>Answer: Circle (x² + y² = constant)</strong></p>
Correct Answer: A