Straight Lines
Distance and position on a line
Grade 11
Question:
<p>A straight line through origin <i>O</i> meets the lines <i>y</i> = <i>x</i> + 3 and <i>y</i> = <i>x</i> + 6 at <i>B</i> and <i>C</i> respectively. A point <i>A</i> is taken on the line <i>y</i> = <i>x</i> such that <i>OA</i><sup>2</sup> = <i>OB</i> · <i>OC</i>. Find the possible positions of vertex <i>C</i> if <i>A</i> ≡ (<i>h</i>, <i>k</i>). (From figure context: Triangle with vertices at <i>B</i>(0,3), <i>A</i>(4,0), <i>D</i>(2, 3/2), and <i>C</i>.) Given \(AB = 5\), \(D \equiv \left(2, \frac{3}{2}\right)\), \(CD = \frac{5\sqrt{3}}{2}\), slope of <i>AB</i> is \(-\frac{3}{4}\), slope of <i>CD</i> is \(\frac{4}{3}\). If \(C \equiv (h, k)\), then \[\frac{h-2}{3/5} = \frac{k - 3/2}{4/5} = \pm \frac{5\sqrt{3}}{2}\] Which of the following are correct values of \(h\) and \(k\)?</p>
<p>\(h = 2\left(1 - \dfrac{3\sqrt{3}}{4}\right),\ k = \dfrac{3}{2}\left(1 - \dfrac{4}{\sqrt{3}}\right)\)</p>
<p>\(h = 2\left(1 + \dfrac{3\sqrt{3}}{4}\right),\ k = \dfrac{3}{2}\left(1 + \dfrac{4}{\sqrt{3}}\right)\)</p>
<p>\(h = 2,\ k = \dfrac{3}{2}\)</p>
<p>\(h = 4,\ k = 3\)</p>
Step-by-Step Solution
Key Concept: Use the parametric form of a line through origin meeting two parallel lines, apply the constraint OA² = OB·OC using distances, and recognize that C lies on a line perpendicular to AB at distance 5√3/2 from D.
<p><strong>Step 1: Parametrize line through origin</strong></p><p>Let the line through O have slope m: y = mx. This meets y = x + 3 at B and y = x + 6 at C.</p><p>At B: mx = x + 3 ⟹ B = (3/(m-1), 3m/(m-1))</p><p>At C: mx = x + 6 ⟹ C = (6/(m-1), 6m/(m-1))</p><p><strong>Step 2: Apply constraint OA² = OB·OC</strong></p><p>OB = |3√(m²+1)|/|m-1|, OC = |6√(m²+1)|/|m-1|</p><p>OB·OC = 18(m²+1)/(m-1)²</p><p>Since A is on y = x with OA² = OB·OC: 2h² = 18(m²+1)/(m-1)²</p><p><strong>Step 3: Geometric insight from given data</strong></p><p>From the figure, D = (2, 3/2) is on line BC. The slope of AB = -3/4 and slope of CD = 4/3.</p><p>Since (-3/4) × (4/3) = -1, lines AB and CD are perpendicular.</p><p><strong>Step 4: Use parametric form for C</strong></p><p>C lies on a line through D perpendicular to AB, at distance 5√3/2 from D:</p><p>The unit direction perpendicular to AB has components (4/5, 3/5)</p><p>Thus: h - 2 = ±(5√3/2)·(3/5) = ±(3√3/2)</p><p>And: k - 3/2 = ±(5√3/2)·(4/5) = ±2√3</p><p><strong>Step 5: Calculate both positions</strong></p><p>For the + sign: h = 2 + 3√3/2, k = 3/2 + 2√3</p><p>For the - sign: h = 2 - 3√3/2, k = 3/2 - 2√3</p><p>∴ Answer: A, B (both solutions with opposite signs)</p>
Correct Answer: A, B