TIFR CSE 2014 | Part A | Question: 13

Question

Let $L$ be a line on the two dimensional plane. $L'$s intercepts with the $X$ and $Y$ axes are respectively $a$ and $b$. After rotating the co-ordinate system (and leaving $L$ untouched), the new intercepts are $a'$ and $b'$ respectively. Which of the following is TRUE?

• A. $\frac{1}{a}+\frac{1}{b}=\frac{1}{a}+\frac{1}{b}$.
• B. $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{a'^{2}}+\frac{1}{b'^{2}}$.
• C. $\frac{b}{a^{2}}+\frac{a}{b^{2}}=\frac{b'}{a'^{2}}+\frac{a}{b'^{2}}$.
• D. $\frac{b}{a}+\frac{a}{b}=\frac{b'}{a'}+\frac{a'}{b'}$.
• E. None of the above.

Comments

if we rotate coordinates by θ , then our a' and b' would be like

a'=acosθ+bsinθ

b'=-asinθ+bcosθ

as i'm able to see , no options is matching after placing values.

https://en.wikipedia.org/wiki/Rotation_of_axes

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No simplification?

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I think answer should be (b). I have added the answer.

Answer 1

If $X-$intercept is $a$ and $Y-$intercept is $b$ then equation of line on $X-Y$ plane is

$$\frac{x}{a} + \frac{y}{b} = 1 \quad \to (1)$$

We can prove it by considering two points on the line, $(a,0)$ as $(x_1,y_1)$ and $(0,b)$ as $(x_2,y_2)$

So, equation of line will be : $y - y_1 = \left ( \frac{y_2-y_1}{x_2-x_1} \right )(x-x_1)$

$\Rightarrow (y - 0) = \left ( \frac{b-0}{0-a} \right )(x-a)$

$\Rightarrow -ay = bx - ab$

On dividing by $ab$ and rearranging,

$\Rightarrow$ $\frac{x}{a} + \frac{y}{b} = 1$

Now, if a point $(x,y)$ lies on the $X-Y$ plane and if we rotate the co-ordinate system by $\theta$ degree then this point becomes $(x',y')$ on new co-ordinate system and relationship is given by :

$\begin{bmatrix} \cos\theta & \sin\theta \\  - \sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} x'\\y' \end{bmatrix}$

So,

• $x' = x\cos\theta +y\sin\theta$ and
• $y'= - x\sin\theta + y\cos\theta$

Now, on the new co-ordinate system, equation on line will be :

$\frac{x'}{a'} + \frac{y'}{b'} = 1$

$\Rightarrow$ $\frac{ x\cos\theta +y\sin\theta}{a'} + \frac{-x\sin\theta + y\cos\theta}{b'} = 1$

$\Rightarrow x\left ( \frac{\cos\theta}{a'} - \frac{\sin\theta}{b'} \right ) + y\left ( \frac{\sin\theta}{a'} + \frac{\cos\theta}{b'} \right ) = 1$

On comparing with equation $(1),$

$\left ( \frac{\cos\theta}{a'} + \frac{-\sin\theta}{b'} \right ) = \frac{1}{a}$ and

$\left ( \frac{\sin\theta}{a'} + \frac{\cos\theta}{b'} \right ) = \frac{1}{b}$

Now, on squaring both sides and adding above $2$ equations :-

$\left ( \frac{1}{a^2} + \frac{1}{b^2} \right ) = \left ( \frac{\cos\theta}{a'} + \frac{-\sin\theta}{b'} \right )^{2} + \left ( \frac{\sin\theta}{a'} + \frac{\cos\theta}{b'} \right )^2$

$ \Rightarrow\left ( \frac{1}{a^2} + \frac{1}{b^2} \right ) = $ $\frac{\cos^{2}\theta}{a'^2} + \frac{\sin^2\theta}{b'^2} -  \frac{2\sin\theta \cos\theta}{a'b'} + \frac{\sin^2\theta}{a'^2} + \frac{\cos^2\theta}{b'^2} + \frac{2\sin\theta \cos\theta}{a'b'}$

$ \Rightarrow \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{a'^2} + \frac{1}{b'^2}$

So, Answer is (b)

Comments

how u taken matrix

$\begin{bmatrix} cos\Theta & -sin\Theta \\ sin\Theta & cos\Theta \end{bmatrix}$??

approach is ok.  but is matrix is general or u rotate it right direction.

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mam,  I did small mistake.. It is $\begin{bmatrix} cos\theta &sin\theta \\ -sin\theta&cos\theta \end{bmatrix}$ not $\begin{bmatrix} cos\theta &-sin\theta \\ sin\theta&cos\theta \end{bmatrix}$

1st matrix rotates the axis and gives the co-ordinates of the same point with respect to new co-ordinate system.. 

2nd matrix rotates a point in the same co-ordinate system. It is called rotation matrix.

in both cases, rotation is anticlockwise..  

proof of 1st matrix is given in above Wikipedia link of 1st comment. 

I guess proof of 2nd matrix is given in book of Gilbert Strang. 

you can find it here also

 https://www.khanacademy.org/partner-content/pixar/sets/rotation/v/sets-8

https://www.khanacademy.org/partner-content/pixar/sets/rotation/v/sets-9

if anything is wrong in answer, you can edit it. 

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As per your explanation

x’=xcos(theta)+ysin(theta)

y’=ycos(theta)-xsin(theta)

Now the intercept wrt to new axis is

(a’,0) which is nothing but (acos(theta),-asin(theta)) as per above transform

Now computing distance to origin with the above we get

a’=a

similarly

b’=b

What Im saying is distances of intercepts remains unchanged by rotating axis.

So all options seem to be correct here

There is something wrong

Can any body help

@arjun sir?

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What Im saying is distances of intercepts remains unchanged by rotating axis.

can you please give an example ?

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Even in the current problem.You can see in the approach you have given.

Just rotate point a,0  by the rotation matrix and calculate distance.

See by performing rotation relative distances dont change.

If the distance from origin before rotation was ‘a’ then after performing rotation still the distance from origin doesnt change in the new system.

Even the length of line doesnt change.Only coordinates change.

Accordingly the oly relation I see is the following

Length of line before rotation

Sqrt(a^2+b^2)

Length of line after rotation

Sqrt(a’^2+b’^2)

Both of above are equal becauseits the same line just rotated.

Equate both and you get

a^2+b^2=a’^2+b’^2

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Just rotate point a,0  by the rotation matrix and calculate distance.

we are not rotating point here. 

(a’,0) which is nothing but (acos(theta),-asin(theta)) as per above transform

why ?

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Agreed we are not rotating point.We are rotating the axis and in the process the points of the line gets transformed to the new axis.

Now a,0 is a point that also gets rotated in this process.Transform this to new axis as per tge rotation matrix. You get the coordinates of the new intercept with respect to old axis as (acos(theta),-asin(theta)).However  relative to new axis it is (a’,0).

But distance from origin is still same in both systems.

Hence a’=a

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You get the coordinates of the new intercept with respect to old axis as (acos(theta),-asin(theta)).
 

it should be $(acos\theta,asin\theta)$. 

However  relative to new axis it is (a’,0)

can you prove new point after rotation i.e. $(acos\theta,asin\theta)$ is same as $(a',0)$ for the new axes ?

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@ankitgupta.1729

In rotation points are like

$x'=x \cos\theta -y \sin\theta$

and $y'=x \sin\theta -y \cos\theta$

right?

Then, matrix should be

$\begin{bmatrix} \cos\Theta &-\sin\Theta \\ \sin\Theta& -\cos\Theta \end{bmatrix}$

Why u have taken different one?

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mam, why  $x'=xcos\theta - ysin\theta$ and $y' = xsin\theta -  ycos\theta \;?$

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@ankitgupta.1729

It is general equation

right?

I want to why u changed it?

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@srestha mam, Please see all the comments and links.

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Good answer, 
In exam however, if comparing doesn't strike, we can also find $a$ in terms of $a'$ & $b'$by substituting the value of $(a,0)$ into the line $\frac{xcos\theta - ysin\theta}{a'}+\frac{xsin\theta + ycos\theta}{b'} = 1$

Also, we can find $b$ in terms of $a'$ & $b'$by substituting the value of $(0,b)$ into the line.

One last clarification, your rotation matrix transforms coordinates clockwise. Also, the rotation doesn't matter clockwise or anticlockwise, as the resulting property still holds.