Question

The point on the curve $y=x^2-3x+2$ at which the tangent is perpendicular to the line $y=x$ is -

• A. $(0,2)$
• B. $(1,0)$
• C. $(-1,6)$
• D. $(2,-2)$

Answer 1

Answer: (B)

$(1,0)$

Let the point be $P(a,b)$
Now the given curve is $y=x^2-3x+2$
Differentiating w.r.t $x$
$\frac{dy}{dx}=2x-3$
Thus slope of tangent at P is $={\left(\frac{dy}{dx}\right)}_{(a,b)}=2a-3$
But given the tangent is perpendicular to line $y=x$
$\Rightarrow 2a-3=-1\Rightarrow a=1$
Also the point P lies on the given curve,
$b=a^2-3a+2=0$
Therefore, the point P is $(1,0)$
Hence, option 'B' is correct.