#### Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 15 Maths Textbook Solution.

Point: $\left ( 1,0 \right )$

Hint:

1. First we will find the slope of line $x-y= 0$ then evaluate with the slope of curve
2. If tangent is perpendicular to the line then  $m_{1} \cdot m_{2}=-1$

Given:

Given curve  $y=x^{2}-3 x+2$  where the tangent is perpendicular to the line  $x-y= 0$

To find:

We have to find the co-ordinate of the point on the given curve.

Solution:

We have,

\begin{aligned} & x-y=0 \\ \Rightarrow \quad & y=x \end{aligned}                                                                                                                                                             … (i)

Comparing equation (i) with equation  $y=m x+c$

$\Rightarrow \quad m_{1}=1$

Again, if the line is perpendicular to the tangent then

\begin{aligned} &\Rightarrow \quad m_{1} \cdot m_{2}=-1 \\ &\Rightarrow \quad 1 \cdot m_{2}=-1 \end{aligned}                                                                                                                                          $\left[\because m_{1}=1\right]$

$\Rightarrow \quad m_{2}=-1$

Given curve,

$y=x^{2}-3 x+2$                                                                                                                                                  … (ii)

On differentiating both side with respect to $x$, we get

$\Rightarrow \quad \frac{d y}{d x}=2 x-3$                                                                                                                        $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

Here slope of tangent is $-1$

\begin{aligned} &\Rightarrow \quad 2 x-3=-1 \\ &\Rightarrow \quad 2 x=2 \\ &\Rightarrow \quad x=1 \end{aligned}

Substituting the value of $x$ in equation (ii), we get

\begin{aligned} &y=1-3+2 \\ &y=0 \end{aligned}

Hence the point is $\left ( x,y \right )= \left ( 1,0 \right )$