Question

The point on the curve y = x² − 3x + 2 where tangent is perpendicular to y = x is

(a) (0, 2)
(b) (1, 0)
(c) (−1, 6)
(d) (2, −2)

Answer 1

(b) (1, 0)

y = x
$\Rightarrow \frac{dy}{dx}=1$

$$\begin{gathered} \text{Let}\left(x_1,y_1\right)\text{be the required point.} \\ \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}, \\ \mathrm{Hence},y_1={x_1}^2-3x_1+2 \\ \mathrm{Now},y=x^2-3x+2 \\ \therefore \frac{dy}{dx}=2x-3 \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=2x_1-3 \end{gathered}$$
The tangent is perpendicular to this line.
∴Slope of the tangent = $\frac{-1}{\text{Slope of the line}}=\frac{-1}{1}=-1$

Now,
$$\begin{gathered} 2x_1-3=-1 \\ \Rightarrow 2x_1=2 \\ \Rightarrow x_1=1 \\ \mathrm{and} \\ {y}_1={x_1}^2-3x_1+2=1-3+2=0 \\ \therefore \left(x_1,y_1\right)=\left(1,0\right) \end{gathered}$$