Question

point a(9,12) rotates around the orgin o in the plane through 60degree in the anticlockwise direction to a new position b find the coordinates of the point​

Answer 1

Let B(h, k) be the new position of the point A(9, 12) when it's rotated about O(0, 0) in anticlockwise direction.

As the point is rotated about origin, it's distance from the origin is not changed. Hence,

$\longrightarrow OB=OA$

$\longrightarrow\sqrt{h^2+k^2}=\sqrt{9^2+12^2}$

$\longrightarrow h^2+k^2=225\quad\quad\dots(1)$

The angle made by OA with x axis will be,

$\longrightarrow\alpha_1=\tan^{-1}(m_1)$

where $m_1$ is slope of OA, i.e.,

$\longrightarrow\alpha_1=\tan^{-1}\left(\dfrac{12-0}{9-0} \right)$

$\longrightarrow\alpha_1=\tan^{-1}\left(\dfrac{4}{3}\right)$

$\longrightarrow\tan\alpha_1=\dfrac{4}{3}$

Let $\alpha_2$ be the angle made by OB with x axis, which is 60° more than $\alpha_1,$ i.e.,

$\longrightarrow\alpha_2=\alpha_1+60^o$

Then slope of OB will be,

$\longrightarrow m_2=\tan\alpha_2$

$\longrightarrow m_2=\tan(\alpha_1+60^o)$

$\longrightarrow m_2=\dfrac{ \tan\alpha_1+\tan60^o}{1-\tan\alpha_1\tan60^o}$

$\longrightarrow\dfrac{k-0}{h-0}=\dfrac{\dfrac{4}{3}+\sqrt3}{1-\dfrac{4}{3}\sqrt3}$

$\longrightarrow\dfrac{k}{h}=\dfrac{4+3\sqrt3}{3-4\sqrt3}\quad\quad\dots(2)$

Squaring,

$\longrightarrow\dfrac{k^2}{h^2}=\dfrac{43+24\sqrt3}{57-24\sqrt3}$

$\longrightarrow\dfrac{k^2}{h^2}=\dfrac{43+24\sqrt3}{100-\left(43+24\sqrt3\right)}$

$\Longrightarrow\dfrac{k^2}{h^2+k^2}=\dfrac{43+24\sqrt3}{100}$

From (1),

$\longrightarrow\dfrac{k^2}{225}=\dfrac{43+24\sqrt3}{100}$

$\longrightarrow k^2=\dfrac{225}{100}(43+24\sqrt3)$

$\longrightarrow k=\dfrac{3}{2}(4+3\sqrt3)$

And from (2),

$\longrightarrow h=\dfrac{3-4\sqrt3}{4+3\sqrt3}\,k$

$\longrightarrow h=\dfrac{3}{2}(3-4\sqrt3)$

Hence the new position is $\bf{\left(\dfrac{3}{2}\left(3-4\sqrt3\right),\ \dfrac{3}{2}\left(4+3\sqrt3\right)\right).}$