Question

The line joining two points A(2,0); B(3,1) is rotated about A in the anticlockwise direction through an angle of 15°. The equation of the line in the new position is:
(a)$x-\sqrt{3}y-2=0$
(b)$x-2y-2=0$
(c)$\sqrt{3}x-y-2\sqrt{3}=0$
(d)None

Answer 1

Hint: Use the formula of$\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}{{m}_{1}}}$, where θ is the angle between two lines and m1 and m2 is the respective slope of the two lines. From this formula, we will get the slope of the required line. And we know that, equation of a line has a slope and a point through which the line is passing. So, we get the slope from the formula and a point “A” is passing through the line after rotation.

Complete step-by-step answer:
Now, m1 is the slope of the line passing through point A and point B.

${{m}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$

Here, point A(x1,y1) and point B(x2,y2). Now, we are substituting the values of x1,y1,x2,y2 in the above equation and we get

${{m}_{1}}=\dfrac{1-0}{3-2}$

$\Rightarrow {{m}_{1}}=1$

Now, we need to find the slope of the line after rotation by the formula I mentioned in the hint.

$\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}{{m}_{1}}}$

Here the value of θ is 15° and m1 we have already calculated. So, substituting these values in above formula we get,

$\tan 15=\dfrac{{{m}_{2}}-1}{1+{{m}_{2}}\left( 1 \right)}$

We know that tan15° = 2-√3 so rewriting the above equation,

$2-\sqrt{3}=\dfrac{{{m}_{2}}-1}{1+{{m}_{2}}}$

Cross multiplying both the sides we get,

$\begin{align}

  & \left( 2-\sqrt{3} \right)\left( 1+{{m}_{2}} \right)={{m}_{2}}-1 \\

 & \Rightarrow 2+2{{m}_{2}}-\sqrt{3}-\sqrt{3}{{m}_{2}}={{m}_{2}}-1 \\

\end{align}$

Now, separating numerals and m2 we get:

$\begin{align}

  & 3-\sqrt{3}=-{{m}_{2}}+\sqrt{3}{{m}_{2}} \\

 & \Rightarrow \sqrt{3}\left( \sqrt{3}-1 \right)={{m}_{2}}\left( \sqrt{3}-1 \right) \\

\end{align}$

In the above expression √3-1 will be cancelled out from both the sides and we get,

$\sqrt{3}={{m}_{2}}$

As we have got m2 and point A is given, we can write the equation of a line as:

$\begin{align}

  & y-0=\sqrt{3}\left( x-2 \right) \\

 & \Rightarrow y=\sqrt{3}x-2\sqrt{3} \\

 & \Rightarrow \sqrt{3}x-y-2\sqrt{3}=0 \\

\end{align}$

Hence, the correct answer is option (c).


Note: The blunder that will be committed in this problem is by taking m2 = 1. As in the formula of $\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}{{m}_{1}}}$, you might be confused what values do m1 and m2 will take. So, always keep in mind that m2 is the slope of the line after rotation and m1 is the slope before rotation.