Straight Lines
Collinearity and locus
Grade 11
Question:
<p>Since the given point lies on the line \(lx + my + n = 0\), so \(a, b, c\) are the roots of the equation \(l\left(\frac{t^3}{t-1}\right) + m\left(\frac{t^2-3}{t-1}\right) + n = 0\), or \(lt^3 + mt^2 + nt - (3m+n) = 0\). Then which of the following are correct?</p>
<p>(a) \(a + b + c = -\frac{m}{l}\)</p>
<p>(b) \(abc = \frac{3m+n}{l}\)</p>
<p>(c) \(ab + bc + ca = \frac{n}{l}\)</p>
<p>(d) \(a + b + c + abc = \frac{3m}{l}\)</p>
Step-by-Step Solution
Key Concept: If a point (x(t), y(t)) lies on the fixed line lx + my + n = 0 for multiple values of t, then substituting the parametric forms yields a polynomial equation whose roots are exactly those t-values. The coefficients of this polynomial encode the line's equation.
<p><strong>Step 1:</strong> Given that point <strong>P(t)</strong> with coordinates <strong>x = t³/(t-1)</strong> and <strong>y = (t²-3)/(t-1)</strong> lies on the fixed line <strong>lx + my + n = 0</strong>.</p><p><strong>Step 2:</strong> Substitute the parametric equations into the line equation:<br/>l·(t³/(t-1)) + m·((t²-3)/(t-1)) + n = 0</p><p><strong>Step 3:</strong> Multiply through by (t-1) to clear denominators:<br/>lt³ + m(t²-3) + n(t-1) = 0<br/>lt³ + mt² + nt - 3m - n = 0</p><p><strong>Step 4:</strong> This cubic equation in t has roots <strong>a, b, c</strong> (the parameter values where the curve intersects the line). By Vieta's formulas:<br/>• Sum of roots: <strong>a + b + c = -m/l</strong><br/>• Sum of products of pairs: <strong>ab + bc + ca = n/l</strong><br/>• Product of roots: <strong>abc = (3m+n)/l</strong></p><p><strong>Step 5:</strong> The correct options (a), (b), and (d) are those matching these Vieta's relations for the polynomial <strong>lt³ + mt² + nt - (3m+n) = 0</strong>.</p><p>∴ Answer: (a), (b), and (d)</p>
Correct Answer: (a), (b), and (d)