Question

The line joining two points A(2, 0) and B(3, 1) is rotated about A in anticlockwise direction throughan angle $\text{1}{\text{5}}^{\text{o}}$ if B goes to C, then C is

• A. $\left(\frac{1}{2}(4+\sqrt{2},\sqrt{6})\right)$
• B. $\left(\frac{1}{2}(6+\sqrt{2},\frac{1}{2}\sqrt{6})\right)$
• C. $\left(\frac{1}{2}(2+\sqrt{2},\frac{1}{2}\sqrt{6})\right)$
• D. $\left(\frac{1}{2}(4+\sqrt{2},\frac{1}{2}\sqrt{6})\right)$

Answer 1

The correct option is A $\left(\frac{1}{2}(4+\sqrt{2},\sqrt{6})\right)$

REF.Image.

Slope of the line AB $=\frac{y_2-y_1}{x_2-x_1}$

$\Rightarrow \frac{1-0}{3-2}=1$

This the angle made with x-axis is ${45}^{\circ }$

$\angle CAB={15}^{\circ }$

$\therefore$ The slope of line CA = $tan(45+15)$

$=tan\left(60\right)$

$=\sqrt{3}$

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

$\sqrt{3}=\frac{0-y_1}{2-x_1}$

$\sqrt{3}(2-x_1)=0-y_1$

$2\sqrt{3}-{\sqrt{3x}}_1=y_1$

$2\sqrt{3}=\sqrt{3}{x}_1-y_1$

From the options

Let$y=\frac{1}{2}\sqrt{6}$

$\Rightarrow x_1=\frac{1}{2}\frac{\sqrt{6}}{\sqrt{3}}+2$

$x_1=2+\frac{1}{\sqrt{2}}$

$x_1=\frac{2\sqrt{2}+1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$x_1=\frac{1}{2}[4+\sqrt{2}]$

$\therefore x_1=\frac{1}{2}[4+\sqrt{2}]$

$y_1=\frac{1}{2}\sqrt{6}$