Quadratic Equations
Algebraic relations from collinearity
Grade 11
Question:
<p>If the points \(\left(\frac{a^3}{a-1}, \frac{a^2-3}{a-1}\right)\), \(\left(\frac{b^3}{b-1}, \frac{b^2-3}{b-1}\right)\), \(\left(\frac{c^3}{c-1}, \frac{c^2-3}{c-1}\right)\) lie on \(L: lx + my + n = 0\) where \(a, b, c\) are real numbers different from 1, then</p>
<p>(a) \(a + b + c = -\frac{m}{l}\)</p>
<p>(b) \(abc = \frac{m+n}{l}\)</p>
<p>(c) \(ab + bc + ca = \frac{n}{l}\)</p>
<p>(d) \(abc - (ab + bc + ca) + 3(a + b + c) = 0\)</p>
Step-by-Step Solution
Key Concept: Transform the parametric point expressions into a polynomial equation in the parameter.
<p>Let $x = \frac{a^3}{a-1}$ and $y = \frac{a^2-3}{a-1}$. Simplify: $x = a^2 + a + 1 + \frac{1}{a-1}$ and $y = a + 1 + \frac{-2}{a-1}$. Since all three points lie on the line: $lx + my + n = 0$. Substituting and factoring by $(a-1)$, we get a cubic equation in $a$ whose roots are $a, b, c$. By Vieta's formulas on this cubic: $a + b + c = -\frac{m}{l}$, $ab + bc + ca = \frac{n}{l}$, and $abc - (ab + bc + ca) + 3(a + b + c) = 0$.</p>
Correct Answer: A, C, D