Straight Lines
Locus
Grade 11
Question:
<p>If a variable line drawn through the intersection of the lines \(\dfrac{x}{3} + \dfrac{y}{4} = 1\) and \(\dfrac{x}{4} + \dfrac{y}{3} = 1\) meets the coordinate axes at A and B, (A ≠ B), then the locus of the mid-point of AB is</p>
<p>\(7xy = 6(x + y)\)</p>
<p>\(4(x + y)^2 - 28(x + y) + 49 = 0\)</p>
<p>\(6xy = 7(x + y)\)</p>
<p>\(14(x + y)^2 - 97(x + y) + 168 = 0\)</p>
Step-by-Step Solution
Key Concept: A variable line through the fixed intersection point of two given lines can be expressed using the family of lines formula L₁ + λL₂ = 0. Find the intersection point, then use the intercept form to relate midpoint coordinates to the parameter.
<p><strong>Step 1: Find intersection point of the two lines</strong></p><p>Given lines: L₁: x/3 + y/4 = 1 and L₂: x/4 + y/3 = 1</p><p>Multiply L₁ by 12: 4x + 3y = 12</p><p>Multiply L₂ by 12: 3x + 4y = 12</p><p>Subtract: x - y = 0 ⟹ x = y</p><p>Substituting in 4x + 3y = 12: 7x = 12 ⟹ x = y = 12/7</p><p>Intersection point P(12/7, 12/7)</p><p><strong>Step 2: Equation of variable line through P</strong></p><p>Let the variable line meet x-axis at A(a, 0) and y-axis at B(0, b), where a ≠ 0, b ≠ 0</p><p>Line equation: x/a + y/b = 1</p><p>Since P(12/7, 12/7) lies on this line: (12/7)/a + (12/7)/b = 1</p><p>Therefore: 12/7(1/a + 1/b) = 1 ⟹ 1/a + 1/b = 7/12</p><p><strong>Step 3: Find locus of midpoint M(h, k)</strong></p><p>Midpoint of AB: h = a/2, k = b/2</p><p>So a = 2h, b = 2k</p><p>Substituting in 1/a + 1/b = 7/12:</p><p>1/(2h) + 1/(2k) = 7/12</p><p>1/h + 1/k = 7/6</p><p><strong>Step 4: Write final locus equation</strong></p><p>Replace h, k with x, y: <strong>1/x + 1/y = 7/6</strong></p><p>Or equivalently: <strong>6(x + y) = 7xy</strong></p><p>∴ Answer: C</p>
Correct Answer: C