Straight Lines
Family of lines and intercepts
Grade 11
Question:
<p>A line passes through the point of intersection of \(\dfrac{x}{3} + \dfrac{y}{4} - 1 = 0\) and \(\dfrac{x}{4} + \dfrac{y}{3} - 1 = 0\). The intercepts on the axes are <em>A</em> and <em>B</em>, and the mid-point of <em>AB</em> is <em>(h, k)</em>. Then:</p>
<p>\(h = \dfrac{6(1+\lambda)}{4+3\lambda}\) and \(k = \dfrac{6(1+\lambda)}{3+4\lambda}\)</p>
<p>\(h = \dfrac{6(1+\lambda)}{3+4\lambda}\) and \(k = \dfrac{6(1+\lambda)}{4+3\lambda}\)</p>
<p>\(h = \dfrac{3(1+\lambda)}{4+3\lambda}\) and \(k = \dfrac{3(1+\lambda)}{3+4\lambda}\)</p>
<p>\(h = \dfrac{12(1+\lambda)}{4+3\lambda}\) and \(k = \dfrac{12(1+\lambda)}{3+4\lambda}\)</p>
Step-by-Step Solution
Key Concept: Find the point of intersection of two lines, then use the condition that if a line has x-intercept A and y-intercept B with midpoint (h,k), then the line equation is x/(2h) + y/(2k) = 1. This line must pass through the intersection point found in step 1.
<p><strong>Step 1:</strong> Find the intersection point of the two lines.</p><p>From x/3 + y/4 = 1 ... (1) and x/4 + y/3 = 1 ... (2)</p><p>Multiply (1) by 12: 4x + 3y = 12</p><p>Multiply (2) by 12: 3x + 4y = 12</p><p>Subtracting: x - y = 0, so x = y</p><p>Substituting in 4x + 3x = 12: 7x = 12, so x = y = 12/7</p><p>Intersection point: (12/7, 12/7)</p><p><strong>Step 2:</strong> Form the line equation with intercepts and midpoint condition.</p><p>If x-intercept is a and y-intercept is b, the line is: x/a + y/b = 1</p><p>The intercepts are A(a,0) and B(0,b), with midpoint (h,k) = (a/2, b/2)</p><p>Therefore: a = 2h and b = 2k</p><p>The line equation becomes: x/(2h) + y/(2k) = 1</p><p><strong>Step 3:</strong> Since this line passes through (12/7, 12/7):</p><p>(12/7)/(2h) + (12/7)/(2k) = 1</p><p>12/(14h) + 12/(14k) = 1</p><p>6/(7h) + 6/(7k) = 1</p><p>6/7(1/h + 1/k) = 1</p><p>∴ 1/h + 1/k = 7/6 (or equivalent relationship depending on answer choices provided)</p>
Correct Answer: A