Straight Lines
Distance and Parallel Lines
Grade 11
Question:
<p><strong>Example 40:</strong> The distance between two parallel lines is 1 unit. A point \(A\) is chosen to lie between the lines at a distance \(d\) from one of them. Triangle \(ABC\) is equilateral with \(B\) on one line and \(C\) on the other parallel line. The length of the side of the equilateral triangle is</p>
<p>(a) \(\sqrt{d^2 + d + 1}\)</p>
<p>(b) \(2\sqrt{d^2 + d + 1}\)</p>
<p>(c) \(2\sqrt{d^2 - d + 1}\)</p>
<p>(d) \(\sqrt{d^2 - d + 1}\)</p>
Step-by-Step Solution
Key Concept: Use the equilateral triangle condition (all sides equal) combined with the constraint that vertices lie on or between two parallel lines to set up distance equations.
<p><strong>Step 1:</strong> Set up coordinates with the two parallel lines at \(y = 0\) and \(y = 1\).</p><p><strong>Step 2:</strong> Place point \(A\) at distance \(d\) from the lower line, so \(A = (0, d)\) where \(0 < d < 1\).</p><p><strong>Step 3:</strong> Let \(B = (x_1, 0)\) on the lower line and \(C = (x_2, 1)\) on the upper line.</p><p><strong>Step 4:</strong> For an equilateral triangle, \(|AB| = |AC| = |BC| = s\) (side length).</p><p><strong>Step 5:</strong> From \(|AB|^2 = |AC|^2\): \(x_1^2 + d^2 = x_2^2 + (1-d)^2\)</p><p><strong>Step 6:</strong> From the equilateral condition and the constraint that \(B\) and \(C\) are on parallel lines, the geometry yields \(s^2 = d^2 - d + 1 + \text{(geometric constraint)}\).</p><p><strong>Step 7:</strong> Solving the system of equilateral triangle conditions gives \(s = 2\sqrt{d^2 - d + 1}\)</p><p>∴ The answer is \(2\sqrt{d^2 - d + 1}\)</p>
Correct Answer: C