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

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(1, 0) y = x \Rightarrow 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 dy/dx = 2x - 3 \text { Slope of the tangent }= \left( dy/dx \right)_\left( x_1 , y_1 \right) = 2 x_1 - 3 The tangent is perpendicular to this line. $\therefore$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)

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