Solution: To find the point on the line $ y = \frac{3}{4}x + 2 $ closest to $ (0, 0) $, we minimize the square of the distance function:

["Topic: How to Find the Point on the Line $ y = \frac{3}{4}x + 2 $ Closest to the Origin: Minimizing the Square of the Distance Function", "Finding the point on a straight line closest to a given point is a fundamental geometric problem with applications in optimization, physics, and computer graphics. In this article, we explore how to determine the closest point on the line $ y = \frac{3}{4}x + 2 $ to the origin $ (0, 0) $ by minimizing the square of the distance function—a method that simplifies calculations while preserving accuracy.", "---", "### The Problem: Closest Point on a Line to the Origin", "Suppose we are given a line defined by $ y = \frac{3}{4}x + 2 $, and we want the coordinates $ (x, y) $ on this line that is nearest to the origin $ (0, 0) $. The direct distance $ d $ from the origin to a point $ (x, y) $ is:", "$$\nd = \sqrt{x^2 + y^2}\n$$", "To minimize $ d $, it is sufficient and more efficient to minimize $ d^2 = x^2 + y^2 $. This avoids the complexity of square roots while yielding the same optimal point.", "---", "### Substituting the Line Equation", "Since $ (x, y) $ lies on the line, we substitute $ y = \frac{3}{4}x + 2 $ into the distance squared function:", "$$\nd^2 = x^2 + \left( \frac{3}{4}x + 2 \right)^2\n$$", "Now expand the expression:", "$$\nd^2 = x^2 + \left( \frac{9}{16}x^2 + 2 \cdot \frac{3}{4}x \cdot 2 + 4 \right)\n= x^2 + \frac{9}{16}x^2 + 3x + 4\n$$", "Combine like terms:", "$$\nd^2 = \left(1 + \frac{9}{16}\right)x^2 + 3x + 4 = \frac{25}{16}x^2 + 3x + 4\n$$", "---", "### Minimizing the Quadratic Function", "The function $ d^2 = \frac{25}{16}x^2 + 3x + 4 $ is a quadratic in standard form $ ax^2 + bx + c $ with $ a = \frac{25}{16} > 0 $, meaning the parabola opens upwards and has a unique minimum at its vertex.", "The $ x $-coordinate of the vertex is given by:", "$$\nx = -\frac{b}{2a} = -\frac{3}{2 \cdot \frac{25}{16}} = -\frac{3}{\frac{50}{16}} = -\frac{3 \cdot 16}{50} = -\frac{48}{50} = -\frac{24}{25}\n$$", "---", "### Finding the Corresponding $ y $-coordinate", "Now substitute $ x = -\frac{24}{25} $ into the line equation $ y = \frac{3}{4}x + 2 $:", "$$\ny = \frac{3}{4} \cdot \left( -\frac{24}{25} \right) + 2 = -\frac{72}{100} + 2 = -\frac{18}{25} + 2 = \frac{-18 + 50}{25} = \frac{32}{25}\n$$", "---", "### Final Answer: The Closest Point", "Thus, the point on the line $ y = \frac{3}{4}x + 2 $ closest to $ (0, 0) $ is:", "$$\n\left( -\frac{24}{25},\ \frac{32}{25} \right)\n$$", "---", "### Why This Method Works", "Minimizing the square of the distance function avoids complications caused by square roots, making differentiation straightforward. The resulting quadratic equation has a unique minimum due to its positive leading coefficient, ensuring the vertex represents the closest point.", "This technique applies broadly to projecting any point onto a line, providing a powerful analytical tool in geometry and optimization.", "---", "### Summary", "To find the closest point on $ y = \frac{3}{4}x + 2 $ to the origin:\n- Substitute the line equation into $ x^2 + y^2 $\n- Form a quadratic in $ x $\n- Minimize using the vertex formula\n- Compute corresponding $ y $", "The solution $ \left( -\frac{24}{25},\ \frac{32}{25} \right) $ is the point on the line nearest to $ (0, 0) $ — efficient, elegant, and grounded in calculus.", "Keywords: closest point on a line, minimize distance function, square of distance, $ y = \frac{3}{4}x + 2 $, origin projection, calculus optimization, coordinate geometry."]