Solution: The closest point occurs where the vector from $ (5, 7) $ to the line is perpendicular to the direction vector $ \beginpmatrix 3 \\ 4 \endpmatrix $. Let the point on the line be $ (2 + 3t, -1 + 4t) $. The vector from $ (5, 7) $ to this point is $ \beginpmatrix 2 + 3t - 5 \\ -1 + 4t - 7 \endpmatrix = \beginpmatrix -3 + 3t \\ -8 + 4t \endpmatrix $. Dot product with $ \beginpmatrix 3 \\ 4 \endpmatrix $ must be zero: $ 3(-3 + 3t) + 4(-8 + 4t) = 0 $. Simplify: $ -9 + 9t - 32 + 16t = 0 \Rightarrow 25t - 41 = 0 \Rightarrow t = \frac4125 $. Substitute back: $ x = 2 + 3 \cdot \frac4125 = \frac17325 $, $ y = -1 + 4 \cdot \frac4125 = \frac15625 $. - Abu Waleed Tea