Question

The coordinates of the points on the curve $y=x^2-3x+2$ where the tangent is perpendicular to the straight line $y=x$ are

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

Answer 1

The correct option is B

$\left(1,0\right)$

Explanation for the correct option:

Step 1: Finding the slope of tangent

Given that $y=x^2-3x+2$

The tangent is perpendicular to the straight line $y=x$.

Slope of $y=x$ is $1$

The slope of the tangent

$$\begin{gathered} =\frac{-1}{1} \\ =-1 \end{gathered}$$

Step 2: Comparing slope with the derivative form to find value of $\left(x,y\right)$

The equation is $y=x^2-3x+2$.

Differentiating both sides

$\frac{dy}{dx}=2x-3$

now slope of tangent =$-1$

Therefore

$$\begin{gathered} \frac{dy}{dx}=-1 \\ \Rightarrow 2x-3=-1 \\ \Rightarrow 2x=-1+3 \\ \Rightarrow 2x=2 \\ \Rightarrow x=1 \end{gathered}$$

Substituting $x=1$ in the given equation $y=x^2-3x+2$

$$\begin{gathered} y=1^2-3\times 1+2 \\ =1-3+2 \\ =0 \end{gathered}$$

Therefore the point is $\left(1,0\right)$

Hence, option (B) is the correct answer.