<p>If \(x_1, x_2, x_3\) and \(y_1, y_2, y_3\) are both in G.P. with the same common ratio, then the points \((x_1, y_1)\), \((x_2, y_2)\) and \((x_3, y_3)\)</p>
<p>lie on a straight line.</p>
<p>lie on an ellipse.</p>
<p>lie on a circle.</p>
<p>are vertices of a triangle.</p>
Step-by-Step Solution
Key Concept: If two G.P. sequences share the same common ratio r, then consecutive points lie on a line passing through the origin. This is because (x_i, y_i) = (x_1·r^(i-1), y_1·r^(i-1)) = r^(i-1)·(x_1, y_1), making all points scalar multiples of each other.
<p><strong>Step 1:</strong> Express the G.P. sequences.</p><p>Let x₁, x₂, x₃ be in G.P. with common ratio r: x₂ = x₁r, x₃ = x₁r²</p><p>Let y₁, y₂, y₃ be in G.P. with the same common ratio r: y₂ = y₁r, y₃ = y₁r²</p><p><strong>Step 2:</strong> Analyze the coordinates.</p><p>Point 1: (x₁, y₁)</p><p>Point 2: (x₁r, y₁r) = r·(x₁, y₁)</p><p>Point 3: (x₁r², y₁r²) = r²·(x₁, y₁)</p><p><strong>Step 3:</strong> Check collinearity.</p><p>All three points are scalar multiples of (x₁, y₁), meaning they lie on the line y = (y₁/x₁)·x passing through the origin.</p><p>Alternatively, verify: The vectors from Point 1 to Point 2 and Point 1 to Point 3 are (x₁r - x₁, y₁r - y₁) = (x₁(r-1), y₁(r-1)) and (x₁r² - x₁, y₁r² - y₁) = (x₁(r²-1), y₁(r²-1)) = (r+1)·(first vector). These are parallel vectors.</p><p>∴ <strong>The three points are collinear and lie on a straight line passing through the origin.</strong></p>
Correct Answer: A