Straight Lines
Collinearity of points
Grade 11
Question:
<p>If the points \(\left(\frac{a^3}{a-1},\, \frac{a^2-3}{a-1}\right)\), \(\left(\frac{b^3}{b-1},\, \frac{b^2-3}{b-1}\right)\) and \(\left(\frac{c^3}{c-1},\, \frac{c^2-3}{c-1}\right)\), where \(a, b, c\) are different from 1, lie on the line \(lx + my + n = 0\), then</p>
<p>\(a + b + c = -\dfrac{m}{l}\)</p>
<p>\(ab + bc + ca = \dfrac{n}{l}\)</p>
<p>\(abc = \dfrac{(m+n)}{l}\)</p>
<p>\(abc - (bc + ca + ab) + 3(a + b + c) = 0\)</p>
Step-by-Step Solution
Key Concept: Recognize that each point has coordinates of the form (P(t)/(t-1), Q(t)/(t-1)) where P and Q are polynomials. These parametric points lie on a line if and only if the numerators satisfy a linear relation derived from the line equation.
<p><strong>Step 1:</strong> Let P = (x,y) = (t³/(t-1), (t²-3)/(t-1)) for parameter t. Then:</p><p>x(t-1) = t³ ⟹ xt - x = t³</p><p>y(t-1) = t² - 3 ⟹ yt - y = t² - 3</p><p><strong>Step 2:</strong> Rearrange: t³ - xt + x = 0 and t² - yt + y - 3 = 0</p><p>From the line equation lx + my + n = 0, we have: l·(t³/(t-1)) + m·((t²-3)/(t-1)) + n = 0</p><p>This gives: lt³ + m(t² - 3) + n(t - 1) = 0</p><p>⟹ lt³ + mt² + nt - m·3 - n = 0</p><p><strong>Step 3:</strong> Since a, b, c are three distinct values satisfying this cubic equation, and the equation is:</p><p>lt³ + mt² + nt - (3m + n) = 0</p><p><strong>Step 4:</strong> By Vieta's formulas for the three roots a, b, c:</p><p>• Sum of roots: a + b + c = -m/l</p><p>• Sum of products of pairs: ab + bc + ca = n/l</p><p>• Product of roots: abc = (3m + n)/l</p><p><strong>Step 5:</strong> Therefore: <strong>a + b + c = -m/l, ab + bc + ca = n/l, abc = (3m + n)/l</strong></p><p>∴ Answer: A</p>
Correct Answer: A