Solution: The closest point minimizes distance. Parametrize the line as $ (t, 2t + 1) $. The squared distance to $ (3, 4) $ is $ (t - 3)^2 + (2t + 1 - 4)^2 = (t - 3)^2 + (2t - 3)^2 $. Differentiate: $ 2(t - 3) + 8t - 12 = 0 \Rightarrow 10t - 18 = 0 \Rightarrow t = 1.8 $. Substituting, the point is $ (1.8, 4.6) $. oxed{(1.8,\ 4.6)}
["The Closest Point on a Line Minimizes Distance: A Step-by-Step Solution", "When trying to find the point on a straight line that is closest to a given point in the plane, a powerful and elegant approach lies in minimizing the distance using calculus. This method proves not only efficient but also mathematically elegant, revealing the geometric truth with algebraic precision.", "### The Setup", "Suppose we want to find the point on the line defined parametrically as\n$$\n(x, y) = (t, 2t + 1)\n$$\nthat is closest to the fixed point $ (3, 4) $. Because distance calculations involve square roots—which complicate differentiation—we simplify by minimizing the squared distance instead.", "### Step 1: Express the squared distance function", "The squared distance $ D^2 $ between $ (t, 2t+1) $ and $ (3, 4) $ is:\n$$\nD^2(t) = (t - 3)^2 + (2t + 1 - 4)^2 = (t - 3)^2 + (2t - 3)^2\n$$", "Expand both terms:\n$$\n(t - 3)^2 = t^2 - 6t + 9\n$$\n$$\n(2t - 3)^2 = 4t^2 - 12t + 9\n$$", "So overall:\n$$\nD^2(t) = t^2 - 6t + 9 + 4t^2 - 12t + 9 = 5t^2 - 18t + 18\n$$", "### Step 2: Minimize using calculus", "To find the minimum, differentiate $ D^2(t) $ with respect to $ t $:\n$$\n\frac{d}{dt}(D^2) = 10t - 18\n$$", "Set the derivative equal to zero to find critical points:\n$$\n10t - 18 = 0 \Rightarrow t = \frac{18}{10} = 1.8\n$$", "Since the square of a distance function is a parabola opening upwards, this critical point corresponds to a minimum.", "### Step 3: Find the closest point", "Substitute $ t = 1.8 $ back into the parametric line $ (t, 2t + 1) $:\n$$\nx = 1.8,\quad y = 2(1.8) + 1 = 3.6 + 1 = 4.6\n$$\nThus, the closest point is\n$$\n(1.8,\ 4.6)\n$$", "### Why this works", "This technique relies on minimizing Euclidean distance by working with squared distance—avoiding the square root while preserving directionality. Parametrizing the line as $ (t, 2t + 1) $ captures all points along the line efficiently. By differentiating and solving, we pinpoint the unique value of $ t $ that yields shortest distance, offering both computational power and geometric clarity.", "### Final Answer", "Boxed:\n$$\n\boxed{(1.8,\ 4.6)}\n$$", "This point minimizes the distance from $ (3, 4) $ along the line $ y = 2x + 1 $, demonstrating a clean, analytical solution to a fundamental geometric problem."]
