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

Answer: (B)

(1, 0)

Answer 2

Let point is (x₁, y₁) \ y₁ = $x_{1}^{2}$ – 3x₁ + 2 ......(i)

Now y = x² – 3x + 2

$\frac{dy}{dx}$ = 2x – 3 Ž $\left( \frac{\mathbf{dy}}{\mathbf{dx}} \right)_{\mathbf{(}\mathbf{x}_{\mathbf{1}}\mathbf{,}\mathbf{y}_{\mathbf{1}}\mathbf{)}}$ = 2x₁ – 3 ......(ii)

Q Tangent is ^ to y = x & slope of this line is 1

\ slope of tangent will be – 1

\ 2x₁ – 3 = – 1 Ž 2x₁ = 2 Žx₁ = 1

& from (i), y₁ = 1 – 3 + 2 = 0

\ (x₁, y₁) = (1, 0)