The point on the curve

Given that the curve $y=x^{2}-3 x+2$ where tangent is perpendicular to $y=x$

Differentiating both w.r.t. $\mathrm{x}$,

$\frac{\mathrm{dy}}{\mathrm{dx}}=1$ and $\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}-3$

$\because$ the point lies on the curve and line both

Slope of the tangent $=-1$

$\Rightarrow 2 x-3=-1$

$\Rightarrow x=1$

And $y=1-3+2$

$\Rightarrow y=0$

So, the required point is $(1,0)$.