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**Soroban** Hello, ^_^Engineer_Adam^_^!

I don't agree with their answer . . . Is there a typo?

The line $\displaystyle y = x$ has slope $\displaystyle m = 1.$

. . Our normal will also have slope $\displaystyle 1.$

A normal to a curve is perpendicular to the tangent at that point.

. . Hence, the tangent will have slope $\displaystyle -1.$

The slope of a tangent is given by the derivative: .$\displaystyle y' \:=\:-\frac{2}{3}x$

The tangent has slope $\displaystyle -1$ when: .$\displaystyle -\frac{2}{3}x\:=\:-1\;\;\Rightarrow\;\; x = \frac{3}{2}$

. . and: .$\displaystyle y \:=\:2-\frac{1}{2}\left(\frac{3}{2}\right)^2\:=\:\frac{5} {4}$

The normal contains the point $\displaystyle \left(\frac{3}{2},\,\frac{5}{4}\right)$ and has slope $\displaystyle m = 1$.

Its equation is: .$\displaystyle y - \frac{5}{4}\:=\:1\left(x - \frac{3}{2}\right)\quad\Rightarrow\quad 4x - 4y\:=\:1$