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# The line joining two points A (2, 0) and B (3, 1) is rotated about A in anti-clockwise direction through an angle of ${{15}^{\circ }}$ . The equation of the line in the new position, is:A. $\sqrt{3}x-y-2\sqrt{3}=0$B. $x-3\sqrt{y}-2=0$C. $\sqrt{3}x+y-2\sqrt{3}=0$D. $x+\sqrt{3}y-2=0$

Last updated date: 14th Sep 2024
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Hint: The formula to calculate the slope of a line between two points is given as follows
$m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Another formula to calculate the slope of a line is given as follows
$m=\tan \theta$
(Where $\theta$ is the angle measured in the anti-clockwise direction made by the line with the x-axis of which the slope is being calculated)
As the line is rotated anticlockwise keeping the point A as fixed, so the slope of the new line is increased by ${{15}^{\circ }}$.
Hence, in this question, we will first find the slope of the line that joins the two points and then write the slope in terms of tan function. Then we can add the needed rotation and then get the new slope as the rotation is done keeping the point A as fixed and also in the anti-clockwise direction.

As mentioned in the question, we have to find the equation of the new line that is formed when the original line is rotated anticlockwise keeping the point A fixed.
Now, we know that the slope of a line ‘m’ passing through two given points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by the formula:
$m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Thus, we can find the slope of our line using this formula.
Putting the value of points A and B in the formula for slope, we get our slope as:
$m=\dfrac{1-0}{3-2}=1$
Thus, the slope of the line is 1.
We also know that the slope of a line is given by $\tan \theta$ where $\theta$ is the angle made by the line with the x-axis measured in the anticlockwise direction.
Hence, the angle made by this line with the x-axis is given as follows
\begin{align} & \tan \theta =1 \\ & \Rightarrow \theta ={{\tan }^{-1}}\left( 1 \right) \\ & \Rightarrow \theta ={{45}^{\circ }} \\ \end{align}
Now, the line is rotated by ${{15}^{\circ }}$, hence, the angle made by the rotated line will be ${{15}^{\circ }}$ more than the angle made by the line when it was not rotated. Let this angle be $\alpha$.
Thus, the angle $\alpha$ made by the line with the x-axis is given as:
\begin{align} & \alpha ={{45}^{\circ }}+{{15}^{\circ }} \\ & \Rightarrow \alpha ={{60}^{\circ }} \\ \end{align}
Thus, the slope of the new line obtained is given by:
\begin{align} & m=\tan \alpha \\ & m=\tan {{60}^{\circ }} \\ & m=\sqrt{3} \\ \end{align}
Now, we know that when a line has a fixed slope ‘m’ and it passes through a fixed point (h,k), then its equation is given as:
$y-k=m\left( x-h \right)$
Now, we know that our required line has a slope of $\sqrt{3}$ and it passes through the fixed point A given by (2,0).
Hence, the equation of the new line can be written as follows:
\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}
Therefore, the equation of the new line is given as:
$\sqrt{3}x-y-2\sqrt{3}=0$

So, the correct answer is “Option A”.

Note: Remember while measuring the angle made by the line with the x-axis, it is always measured in the anticlockwise direction. If we measure it in the clockwise direction, we should add a minus (-) sign in front of it otherwise it will result in an error in the solution. We also could have used a different general equation to find our required equation given by:
$y=mx+c$
Where, ‘m’ is the slope of the line and ‘c’ is the y-intercept. Here, $m=\sqrt{3}$
Thus, putting this value in the general equation we get:
$y=\sqrt{3}x+c$
But we know that this line also passes through the point A(2,0). Thus, by putting this point in the equation, we will get the value of ‘c’.
It is given as:
\begin{align} & 0=\sqrt{3}.2+c \\ & \Rightarrow c=-2\sqrt{3} \\ \end{align}
Thus, our equation will become:
\begin{align} & y=\sqrt{3}x-2\sqrt{3} \\ & \Rightarrow \sqrt{3}x-y-2\sqrt{3}=0 \\ \end{align}