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: $t^2 - 8t + 16 + \frac{1}{4}t^2 - 2t + 4 = \frac{5}{4}t^2 - 10t + 20$. Taking derivative and setting to zero: $\frac{5}{2}t - 10 = 0 \Rightarrow t = 4$. Substituting back, $y = -\frac{1}{2}(4) + 5 = 3$. Thus, the closest point is $(4, 3)$, which lies on the line. $\boxed{(4, 3)}$

Wed, 22 Apr 2026 11:26:31 +0000

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