Question

The line joins two points$A(2,0)$, and$B(3,1)$ is rotated about A in an anti-clockwise direction through an angle of $15°$. 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. $x+3\sqrt{y}–2=0$

Answer 1

The correct option is A

$\sqrt{3}x–y–2\sqrt{3}=0$

Step 1: The slope of the line with points $A$ and$B$

So, the slope $m$ of the line using point$A(2,0)$and$B(3,1)$

$\Rightarrow m=\frac{y_2-y_1}{x_2-x_1}$

$\Rightarrow m=\frac{\left(1-0\right)}{\left(3-2\right)}=1$

$$\begin{gathered} \Rightarrow m=\tan \left(\theta \right)=1 \\ \Rightarrow \theta =45° \end{gathered}$$

The line is rotated in anticlockwise direction through an angle of $15°$

then, a new angle$=15°+45°=60°$

So, the new slope of the line $=m\text{'}=\tan \left(60°\right)=\sqrt{3}$

Equation of a line with slope $m\text{'}$and point$A(2,0)$ is,

$$\begin{gathered} y-y_1=m\text{'}\left(x-x_1\right) \\ \Rightarrow y-0=\sqrt{3}\left(x-2\right) \\ \Rightarrow y=\sqrt{3}x-2\sqrt{3} \\ \Rightarrow \sqrt{3}x-y-2\sqrt{3}=0 \end{gathered}$$

Hence, the correct option is $\mathrm{Option}\left(\mathrm{A}\right)$.