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

Answers (1)


               Point: \left ( 1,0 \right )


  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 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.


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 )

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