At what points on the curve `y=x^2-4x+5` is the tangent perpendicular to the line `2y+x=7` ?

Answer

Point is `(3,2)`

Explanation

Let `(x_{1}, y_{1})` be required points

slope `=-1 / 2`

slope of line perpendicular `=2`

as `(x_{1}, y_{1})` lies on curve

$$ y_{1}=x_{1}^{2}-4 x_{1}+5 \cdots-(1) $$
`\frac{d y}{d x}=2 x_{1}-4=2`
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