# Point A is at (3 ,7 ) and point B is at (3 ,-4 ). Point A is rotated pi/2  clockwise about the origin. What are the new coordinates of point A and by how much has the distance between points A and B changed?

Feb 20, 2018

New coordinates of point A is $\left(- 7 , 3\right)$ and Distance between A and B changed by $1.1$ unit.

#### Explanation:

$A \left(3 , 7\right) \mathmr{and} B \left(3 , - 4\right)$ . Clockwise rotation of point A is

$\alpha = \frac{\pi}{2} \therefore$ Counterclockwise rotation of point A is

$\theta = 2 \pi - \alpha = 2 \pi - \frac{\pi}{2} = \frac{3 \pi}{2}$. New coordinates of

$A \left(x ' , y '\right)$ can be found by the fomula ,

$x ' = x \cos \theta + y \sin \theta \mathmr{and} y ' = y \cos \theta - x \sin \theta$

$\therefore x ' = 3 \cdot \cos \left(\frac{3 \pi}{2}\right) + 7 \cdot \sin \left(\frac{3 \pi}{2}\right) = 0 + \left(- 7\right) = - 7$

x'= -7; y'= 7* cos((3pi)/2)- 3 * sin((3pi)/2)

$= 7 \cdot 0 - 3 \cdot \left(- 1\right) = 3 \therefore y ' = 3 \therefore \left(x ' , y '\right) = \left(- 7 , 3\right)$

Distance between two points $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$ is

$D = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$ . Orginal distance between

points $A \left(3 , 7\right) \mathmr{and} B \left(3 , - 4\right)$ is

${D}_{o} = \sqrt{{\left(3 - 3\right)}^{2} + {\left(7 + 4\right)}^{2}} = \sqrt{121} = 11.0$ unit.

New distance between points $A \left(- 7 , 3\right) \mathmr{and} B \left(3 , - 4\right)$ is

${D}_{n} = \sqrt{{\left(- 7 - 3\right)}^{2} + {\left(3 + 4\right)}^{2}} = \sqrt{149} \approx 12.2$ unit

Distance between A and B changed by $12.2 - 11.1 = 1.1$ unit .

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