Solution: The closest point is the projection of $(4, 3)$ onto the line. The formula for the projection of a point $(x_0, y_0)$ onto $ax + by + c = 0$ is used. Rewriting the line as $\frac{1}{2}x + y - 5 = 0$, we compute the projection. Alternatively, parametrize the line and minimize distance. Let $x = t$, then $y = -\frac{1}{2}t + 5$. The squared distance to $(4, 3)$ is $(t - 4)^2 + \left(-\frac{1}{2}t + 5 - 3\right)^2 = (t - 4)^2 + \left(-\frac{1}{2}t + 2\right)^2$. Expanding: $t^2 - 8t + 16

["Solution: Finding the Closest Point via Line Projection (Simple & Efficient Method)", "When tasked with finding the point on a line closest to a given point, geometric projection offers a powerful and precise solution. In this article, we’ll explore how to compute the closest point on the line to the point $(4, 3)$ using a direct projection formula — and confirm an elegant alternative via parametrization.", "---", "### Understanding the Line Equation", "The standard form of a line is $ax + by + c = 0$. Given the point $(4, 3)$, we first rewrite the line in this form. The line passing through $(4, 3)$ with slope $-\frac{1}{2}$ has equation:", "$$\ny = -\frac{1}{2}x + 5\n$$", "Rewriting into standard form:", "$$\n\frac{1}{2}x + y - 5 = 0\n$$", "This matches the form $ax + by + c = 0$ with $a = \frac{1}{2}$, $b = 1$, $c = -5$.", "---", "### Projecting a Point Onto a Line", "The closest point on a line to a given point $(x_0, y_0)$ is the orthogonal projection of the point onto the line. For line $ax + by + c = 0$, the projection formula is:", "$$\n(x, y) = \left( x_0 - a \frac{ax_0 + by_0 + c}{a^2 + b^2},\ y_0 - b \frac{ax_0 + by_0 + c}{a^2 + b^2} \right)\n$$", "However, a simpler method uses parametrization to minimize distance directly — which we’ll explore next.", "---", "### Alternative: Parametrize the Line and Minimize Distance", "Let us parametrize the line $y = -\frac{1}{2}x + 5$ by setting $x = t$, so $y = -\frac{1}{2}t + 5$.\nThe point to compare is $(4, 3)$.", "The squared distance $D^2$ from $(t, -\frac{1}{2}t + 5)$ to $(4, 3)$ is:", "$$\nD^2 = (t - 4)^2 + \left(-\frac{1}{2}t + 5 - 3\right)^2\n= (t - 4)^2 + \left(-\frac{1}{2}t + 2\right)^2\n$$", "Expanding:", "$$\n(t - 4)^2 = t^2 - 8t + 16\n$$\n$$\n\left(-\frac{1}{2}t + 2\right)^2 = \frac{1}{4}t^2 - 2t + 4\n$$", "Add both:", "$$\nD^2 = t^2 - 8t + 16 + \frac{1}{4}t^2 - 2t + 4 = \frac{5}{4}t^2 - 10t + 20\n$$", "Take derivative with respect to $t$:", "$$\n\frac{d(D^2)}{dt} = \frac{5}{2}t - 10\n$$", "Set derivative to zero:", "$$\n\frac{5}{2}t - 10 = 0 \Rightarrow t = 4\n$$", "Now substitute $t = 4$ into the parametrization:", "- $x = 4$\n- $y = -\frac{1}{2}(4) + 5 = -2 + 5 = 3$", "So the closest point is $(4, 3)$ — and since it lies on the line, it is the closest point.", "Vertex Insight: The point $(4, 3)$ lies exactly on the line — so no projection is needed beyond recognizing it’s on the line.", "$$\n\boxed{(4, 3)}\n$$", "---", "### Why This Works — Intuition and Verification", "Projection ensures the line segment joining $(4,3)$ and the closest point is perpendicular to the original line — satisfying both geometric and algebraic consistency. Since $(4,3)$ satisfies the line equation:", "$$\n\frac{1}{2}(4) + 3 - 5 = 2 + 3 - 5 = 0\n$$", "it lies on the line. Thus, the shortest distance is zero, and the closest point is $(4, 3)$.", "---", "### Final Summary", "Finding the closest point on a line to a given point can be efficiently solved either by projection formulas or parametrization and minimization. In this case, rewriting the line, parametrizing, and minimizing confirmed that $(4, 3)$ lies directly on the line — making it the closest point.", "Best Practice Tip: When $P$ lies on line $L$, the projection is $P$ itself. Always verify point-line inclusion before applying projection algorithms.", "This approach combines geometry, algebra, and calculus to deliver a clear, accurate, and efficient solution."]