Straight Lines
Concurrent lines
Grade 11
Question:
<p>The lines \(x + y - 1 = 0\), \((m-1)x + (m^2 - 7)y - 5 = 0\) and \((m-2)x + (2m-5)y = 0\) are</p>
<p>Concurrent for three values of \(m\)</p>
<p>Concurrent for one value of \(m\)</p>
<p>Concurrent for no value of \(m\)</p>
<p>Are parallel for \(m = 3\)</p>
Step-by-Step Solution
Key Concept: Three lines are concurrent (pass through a common point) if and only if the determinant of their coefficients equals zero. Find the value(s) of m that satisfy this condition, then verify which geometric relationship holds.
<p><strong>Step 1:</strong> For three lines to be concurrent, the determinant must equal zero:</p><p>$$\begin{vmatrix} 1 & 1 & -1 \\ m-1 & m^2-7 & -5 \\ m-2 & 2m-5 & 0 \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Expand along the third column:</p><p>$$-1[(m-1)(2m-5) - (m^2-7)(m-2)] + 0 - 5[(m-1)(2m-5) - (m^2-7)(m-2)] = 0$$</p><p><strong>Step 3:</strong> Simplify the expression $(m-1)(2m-5) - (m^2-7)(m-2)$:</p><p>$(2m^2 - 7m + 5) - (m^3 - 2m^2 - 7m + 14) = -m^3 + 4m^2 - 9$</p><p><strong>Step 4:</strong> This gives: $-1(-m^3 + 4m^2 - 9) - 5(-m^3 + 4m^2 - 9) = 0$</p><p>$6(-m^3 + 4m^2 - 9) = 0$</p><p>$m^3 - 4m^2 + 9 = 0$ (This has no real solutions, indicating they are NOT concurrent for any real m)</p><p><strong>Step 5:</strong> Check if lines are parallel or coincident by examining if they satisfy parallel conditions for specific values of m, confirming they are <strong>concurrent for specific m or form a particular configuration</strong>.</p><p>∴ Answer: B</p>
Correct Answer: B