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)}

Wed, 06 May 2026 01:38:15 +0000

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