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(D) The given line is $2x - 2y = 3$, which can be written as $y = x - 1.5$. The slope of this line is $m = 1$.Since the normal is parallel to this lin

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$(e^{-2}, -2e^{-2})$

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(D) The given line is $2x - 2y = 3$, which can be written as $y = x - 1.5$. The slope of this line is $m = 1$.Since the normal is parallel to this line, the slope of the normal $m_n = 1$.The slope of the tangent $m_t$ is given by $-1 / m_n = -1 / 1 = -1$.For the curve $y = x \log x$, the derivative is $\frac{dy}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1$.Equating the slope of the tangent to the derivative: $\log x + 1 = -1$, which gives $\log x = -2$.Thus, $x = e^{-2}$.Substituting $x = e^{-2}$ into the curve equation: $y = e^{-2} \log(e^{-2}) = e^{-2} (-2) = -2e^{-2}$.Therefore, the point is $(e^{-2}, -2e^{-2})$.

The coordinates of the point on the curve $y = x \log x$ at which the normal is parallel to the line $2x - 2y = 3$ are:

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