Solution: We minimize the distance from a point \((x, 2x+1)\) on the line to \((4, 3)\). The squared distance is:

["Title: Minimize Distance from a Point on a Line to a Fixed Point: A Step-by-Step Solution Using Squared Distance", "---", "Introduction\nIn geometry, finding the shortest distance from a point to a line often involves minimizing the distance between them. But computing square roots introduces complexity in optimization. A clever solution is to minimize the squared distance—an equivalent method that simplifies calculations without affecting the result.", "This article explores how to minimize the squared distance from a variable point ((x, 2x+1)) on the line described by (y = 2x + 1) to the fixed point ((4, 3)). We’ll derive the formula, apply optimization, and confirm why using squared distance is both efficient and precise.", "---", "### Understanding the Problem", "We are given:\n- A line defined parametrically or implicitly by (y = 2x + 1),\n- A fixed point (P = (4, 3)),\n- A variable point (Q = (x, 2x+1)) lying on the line.", "Our goal is to minimize the squared distance between (Q) and (P), avoiding the square root while retaining mathematical rigor.", "The squared distance (D^2) between points ((x_1, y_1)) and ((x_2, y_2)) is:\n[\nD^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\n]", "For our case:\n[\nD^2 = (x - 4)^2 + \left((2x + 1) - 3\right)^2\n]", "---", "### Step 1: Write the Expression for Squared Distance", "Substitute the coordinates:\n[\nD^2 = (x - 4)^2 + (2x + 1 - 3)^2 \n= (x - 4)^2 + (2x - 2)^2\n]", "Expand both terms:\n[\n(x - 4)^2 = x^2 - 8x + 16\n]\n[\n(2x - 2)^2 = 4x^2 - 8x + 4\n]", "Add them:\n[\nD^2 = x^2 - 8x + 16 + 4x^2 - 8x + 4 = 5x^2 - 16x + 20\n]", "---", "### Step 2: Minimize the Quadratic Function", "The expression (D^2 = 5x^2 - 16x + 20) is a quadratic in (x), opening upwards (since coefficient of (x^2) is positive). Its minimum occurs at the vertex:\n[\nx = -\frac{b}{2a} = -\frac{-16}{2 \cdot 5} = \frac{16}{10} = 1.6\n]", "---", "### Step 3: Find the Closest Point on the Line", "Substitute (x = 1.6) into the line equation:\n[\ny = 2(1.6) + 1 = 3.2 + 1 = 4.2\n]", "Thus, the minimizing point is (Q = (1.6, 4.2)), approximately ((1.6, 4.2)).", "---", "### Step 4: Compute Squared Distance at Minimum", "Plug (x = 1.6) back into (D^2):\n[\nD^2 = 5(1.6)^2 - 16(1.6) + 20 = 5(2.56) - 25.6 + 20 = 12.8 - 25.6 + 20 = 7.2\n]", "So, the minimum squared distance is 7.2.", "---", "### Why Use Squared Distance?", "- Simpler calculus: Minimizing (D^2) avoids square roots, making derivatives easier.\n- Same minimization point: The (x)-value that minimizes (D^2) also minimizes the original distance (D).\n- Efficient computation: Works well for applications like computer graphics, machine learning, and geometric modeling.", "---", "### Conclusion", "By leveraging the squared distance formula, we efficiently minimized the distance from the point ((x, 2x+1)) on the line to ((4, 3)). The method reduces complex square root optimization to a straightforward quadratic minimization—proving that mathematical elegance and computational efficiency go hand in hand.", "---", "Save computation, trust the method: Use squared distance to find the closest point on a line—ideal for geometric optimization in algebra and calculus!", "---", "Keywords: minimum distance, squared distance, line optimization, geometry problem, calculus, algebraic minimization, closest point on line, derivative-free optimization", "---", "Meta Description:\nDiscover how to minimize squared distance from ((x, 2x+1)) to ((4, 3)) using algebra—avoid square roots and simplify distance calculations. Perfect for geometry students and educators!"]