Solution: The closest point occurs where the vector from $ (5, 7) $ to the line is perpendicular to the direction vector $ \begin{pmatrix} 3 \\ 4 \end{pmatrix} $. Let the point on the line be $ (2 + 3t, -1 + 4t) $. The vector from $ (5, 7) $ to this point is $ \begin{pmatrix} 2 + 3t - 5 \\ -1 + 4t - 7 \end{pmatrix} = \begin{pmatrix} -3 + 3t \\ -8 + 4t \end{pmatrix} $. Dot product with $ \begin{pmatrix} 3 \\ 4 \end{pmatrix} $ must be zero: $ 3(-3 + 3t) + 4(-8 + 4t) = 0 $. Simplify: $ -9 + 9t - 32 + 16t = 0 \Rightarrow 25t - 41 = 0 \Rightarrow t = \frac{41}{25} $. Substitute back: $ x = 2 + 3 \cdot \frac{41}{25} = \frac{173}{25} $, $ y = -1 + 4 \cdot \frac{41}{25} = \frac{156}{25} $.

Mon, 04 May 2026 01:48:00 +0000

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