Question

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

Options

• (0, 2)

• (1, 0)

• (−1, 6)

• (2, −2)

Answer 1

(1, 0)

 

`y = x`

\[\Rightarrow \frac{dy}{dx} = 1\]

\[\text { Let }\left( x_1 , y_1 \right)\text { be the required point. }\]

\[\text { Since, the point lies on the curve,} \]

\[\text { Hence }, y_1 = {x_1}^2 - 3 x_1 + 2\]

\[\text { Now }, y = x^2 - 3x + 2\]

\[ \therefore \frac{dy}{dx} = 2x - 3\]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = 2 x_1 - 3\]

The tangent is perpendicular to this line.

∴Slope of the tangent = \[\frac{- 1}{\text { Slope of the line }} = \frac{- 1}{1} = - 1\]

Now,

\[2 x_1 - 3 = - 1\]

\[ \Rightarrow 2 x_1 = 2\]

\[ \Rightarrow x_1 = 1\]

\[\text { and }\]

\[ y_1 = {x_1}^2 - 3 x_1 + 2 = 1 - 3 + 2 = 0\]

\[ \therefore \left( x_1 , y_1 \right) = \left( 1, 0 \right)\]