Question:
The equation of tangent at those points where the curve $y=x^{2}-3 x+2$ meets $x$-axis are
A. $x-y+2=0=x-y-1$
B. $x+y-1=0=x-y-2$
C. $x-y-1=0=x-y$
D. $x-y=0=x+y$
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.