Straight Lines
Collinearity of Points
Grade 11
Question:
<p>Let \(x_1 = x_1,\ x_2 = x_1 r,\ x_3 = x_1 r^2\) and \(y_1 = y_1,\ y_2 = y_1 r,\ y_3 = y_1 r^2\). Which of the following is true about the points \(A \equiv (x_1, y_1)\), \(B \equiv (x_2, y_2)\), \(C \equiv (x_3, y_3)\)?</p>
<p>They form a triangle</p>
<p>They are collinear</p>
<p>They form a right-angled triangle</p>
<p>They form an equilateral triangle</p>
Step-by-Step Solution
Key Concept: Three points with coordinates in geometric progression (with same ratio r) are collinear if and only if the determinant of their position vectors equals zero. This is because collinearity requires the slope between any two pairs to be identical.
<p><strong>Step 1:</strong> Three points A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) are collinear if and only if:</p><p>$$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Substitute x₂ = x₁r, x₃ = x₁r² and y₂ = y₁r, y₃ = y₁r²:</p><p>$$\begin{vmatrix} x_1 & y_1 & 1 \\ x_1r & y_1r & 1 \\ x_1r^2 & y_1r^2 & 1 \end{vmatrix}$$</p><p><strong>Step 3:</strong> Factor out x₁ from column 1 and y₁ from column 2:</p><p>$$x_1y_1\begin{vmatrix} 1 & 1 & 1 \\ r & r & 1 \\ r^2 & r^2 & 1 \end{vmatrix}$$</p><p><strong>Step 4:</strong> R₂ → R₂ - rR₁ and R₃ → R₃ - r²R₁:</p><p>$$x_1y_1\begin{vmatrix} 1 & 1 & 1 \\ 0 & 0 & 1-r \\ 0 & 0 & 1-r^2 \end{vmatrix} = 0$$</p><p><strong>Step 5:</strong> Since columns 1 and 2 are identical in rows 2 and 3, the determinant equals zero for all values of r and non-zero x₁, y₁.</p><p>∴ The three points are <strong>always collinear</strong>, regardless of the value of r. Answer: B</p>
Correct Answer: B