Straight Lines
Family of Lines / Concurrent Lines
Grade 11
Question:
<p>Let \(P\) be any point on the line \(x - y + 3 = 0\) and \(A\) be a fixed point \((3, 4)\). If the family of lines given by the equations \((3\sec\theta + 5\csc\theta)x + (7\sec\theta - 3\csc\theta)y + 11(\sec\theta - \csc\theta) = 0\) are concurrent at a point \(B\) for all permissible value of \(\theta\), then:</p>
<p>(a) sum of the abscissa and ordinate of point \(B\) is equal to \(-1\).</p>
<p>(b) product of the abscissa and ordinate of point \(B\) is equal to \(-2\).</p>
<p>(c) maximum value of \(|PA - PB|\) is \(2\sqrt{10}\).</p>
<p>(d) minimum value of \(PA + PB\) is \(2\sqrt{34}\).</p>
Step-by-Step Solution
Key Concept: A family of lines is concurrent at a fixed point B if we can eliminate the parameter (sec θ and csc θ) by finding a point that satisfies the equation for all values of θ. Rewrite the equation as a linear combination: (3x + 7y + 11)sec θ + (-5x - 3y - 11)csc θ = 0, which holds for all θ only when both coefficients equal zero simultaneously.
<p><strong>Step 1:</strong> Rewrite the given equation grouping sec θ and csc θ terms:</p><p>(3x + 7y + 11)sec θ + (-5x - 3y - 11)csc θ = 0</p><p><strong>Step 2:</strong> For this equation to hold for all permissible values of θ (where both sec θ and csc θ are independent), both coefficients must be zero:</p><p>3x + 7y + 11 = 0 ... (i)</p><p>-5x - 3y - 11 = 0 ... (ii)</p><p><strong>Step 3:</strong> Solve the system. From equation (i): 3x + 7y = -11. From equation (ii): 5x + 3y = -11.</p><p>Multiply (i) by 5 and (ii) by 3: 15x + 35y = -55 and 15x + 9y = -33</p><p>Subtract: 26y = -22, so y = -11/13</p><p>Substitute back: 3x + 7(-11/13) = -11 → 3x = -11 + 77/13 = (-143 + 77)/13 = -66/13 → x = -22/13</p><p>Therefore, B = (-22/13, -11/13)</p><p><strong>Step 4:</strong> Verify geometric relationships:</p><p>• Check if A(3,4) lies on line x - y + 3 = 0: 3 - 4 + 3 = 2 ≠ 0 (A is not on the line)</p><p>• Slope of AB = (-11/13 - 4)/(-22/13 - 3) = (-11/13 - 52/13)/(-22/13 - 39/13) = (-63/13)/(-61/13) = 63/61</p><p>• Slope of line x - y + 3 = 0 is 1</p><p>• Since 63/61 ≠ 1, AB is not parallel to the given line containing P</p><p>• Distance from A to line: |3 - 4 + 3|/√2 = 2/√2 = √2</p><p>• Distance from B to line: |-22/13 + 11/13 + 3|/√2 = |(-22 + 11 + 39)/13|/√2 = 28/(13√2) = 14√2/13</p><p>∴ Answer: ACD</p>
Correct Answer: ACD