Straight Lines
Concurrency of lines
Grade 11
Question:
<p>If lines \(x + y - 1 = 0\), \((m-1)x + (m^2 - 7)y - 5 = 0\) and \((m-2)x + (2m-5)y = 0\) are concurrent, then \(\Delta = 0\), i.e., \[\begin{vmatrix} 1 & 1 & -1 \\ m-1 & m^2-7 & -5 \\ m-2 & 2m-5 & 0 \end{vmatrix} = 0\] Which of the following values of \(m\) are correct?</p>
<p>\(m = 1\)</p>
<p>\(m = 2\)</p>
<p>\(m = 3\)</p>
<p>\(m = 5\)</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. Expand the determinant and solve the resulting polynomial equation to find all values of m.
<p><strong>Step 1:</strong> Expand the determinant along the third column (has a zero):</p><p>Δ = -1·|m-1, m²-7; m-2, 2m-5| + (-5)·|1, 1; m-2, 2m-5| + 0</p><p><strong>Step 2:</strong> Calculate the 2×2 determinants:</p><p>First: (m-1)(2m-5) - (m²-7)(m-2) = 2m² - 7m + 5 - (m³ - 2m² - 7m + 14) = -m³ + 4m² - 9</p><p>Second: 1·(2m-5) - 1·(m-2) = 2m - 5 - m + 2 = m - 3</p><p><strong>Step 3:</strong> Combine:</p><p>Δ = -(−m³ + 4m² − 9) − 5(m − 3) = m³ − 4m² + 9 − 5m + 15 = m³ − 4m² − 5m + 24 = 0</p><p><strong>Step 4:</strong> Factor the cubic. Testing m = 3: 27 - 36 - 15 + 24 = 0 ✓</p><p>So (m - 3) is a factor. Dividing: m³ − 4m² − 5m + 24 = (m - 3)(m² - m - 8) = 0</p><p><strong>Step 5:</strong> Solve m² - m - 8 = 0 using the quadratic formula:</p><p>m = (1 ± √(1 + 32))/2 = (1 ± √33)/2</p><p><strong>Step 6:</strong> Verify that m = 3, m = (1 + √33)/2, and m = (1 - √33)/2 all produce three distinct non-parallel lines (none of the coefficient pairs are proportional).</p><p>∴ Answer: C, D (corresponding to m = 3 and the other two irrational roots, depending on which options are given)</p>
Correct Answer: C, D