Q. The coordinates of the point on the curve $y = x^{2} - 3x + 2\$where the tangent is perpendicular to the straight line $y = x\ are$
A) (0.2)
B) (1.0)
C) $( - 1,6)$
D) (2,$- 2)$
Explanation
(b)
Given curve is
$\ \ \ \ y = x^{2} - 3x + 2$
$\ \ \ \ \frac{dy}{dx} = 2x - 3$
$\ \ \ \ (2x - 3) \times 1 = - 1$
$\Rightarrow \ \ \ \ \ \ 2x - 3 = - 1$
$\Rightarrow \ \ \ \ \ \ \ 2x = 2$
$\Rightarrow \ \ \ \ \ \ \ x = 1$
$\therefore\ \ \ \ \ \ \ \ \ y = 0$
$\Rightarrow Required\ point\ is(1,0).$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.