The equation of tangent

Solution:

Given that the curve $y=x^{2}-3 x+2$

$\Rightarrow \frac{d y}{d x}=2 x-3$

The tangent passes through point $(x, 0)$

$\Rightarrow 0=x^{2}-3 x+2$

$\Rightarrow(x-2)(x-1)=0$

$\Rightarrow x=1$ or 2

Equation of the tangent:

$\left(y-y_{1}\right)=$ Slope of tangent $x\left(x-x_{1}\right)$

Case: 1

When $x=2$

Slope of tangent $=1$

Equation of tangent:

$y=1 \times(x-2)$

$\Rightarrow x-y-2=0$

Case: 2

When $x=1$

Slope of tangent $=-1$

Equation of tangent:

$y=-1 \times(x-1)$

$\Rightarrow x+y-1=0$

Hence, option B is correct.