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Solution

Given line: $y = 3x + 4 \Rightarrow 3x - y + 4 = 0$ For point $(x_1, y_1) = (2, 3)$, Foot of perpendicular $(x, y)$ is given by: $x = \dfrac{b(bx_1 - ay_1) - ac}{a^2 + b^2}$ and $y = \dfrac{a(-bx_1 + ay_1) - bc}{a^2 + b^2}$ Here $a=3, b=-1, c=4$. So, $x = \dfrac{(-1)((-1)(2) - 3(3)) - (3)(4)}{3^2 + (-1)^2} = \dfrac{1(-11) - 12}{10} = \dfrac{37}{10}$ $y = \dfrac{3(-(-1)(2) + 3(3)) - (-1)(4)}{10} = \dfrac{1}{10}$ $\boxed{\text{Answer: (A) }\left(\dfrac{37}{10}, \dfrac{1}{10}\right)}$