# Point A is at (-2 ,-4 ) and point B is at (-3 ,6 ). Point A is rotated (3pi)/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?

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Feb 9, 2018

Distance between A and B changed by $5.93$ unit.

#### Explanation:

$A \left(- 2 , - 4\right) \mathmr{and} B \left(- 3 , 6\right)$ . Clockwise rotation of point A is

$\theta = \frac{3 \pi}{2} \therefore$ Counterclockwise rotation of point A is

$\theta = \left(2 \pi\right) - \frac{3 \pi}{2} = \frac{\pi}{2}$.

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

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

$\therefore x ' = \left(- 2\right) \cdot \cos 90 + \left(- 4\right) \cdot \sin 90 = - 2 \cdot 0 - 4 \cdot 1$

x' = -4 ; y'= -4 * cos 90+ 2 * sin 90 = -4 * 0+2 * 1= 2

Therefore,new coordinates of A are $\left(x ' , y '\right) = \left(- 4 , 2\right)$

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 $A \left(- 2 , - 4\right) \mathmr{and} B \left(- 3 , 6\right)$ is

${D}_{o} = \sqrt{{\left(- 2 + 3\right)}^{2} + {\left(- 4 - 6\right)}^{2}} = \sqrt{101} \approx 10.05$ unit.

New distance between points $A \left(- 4 , 2\right) \mathmr{and} B \left(- 3 , 6\right)$ is

${D}_{n} = \sqrt{{\left(- 4 + 3\right)}^{2} + {\left(2 - 6\right)}^{2}} = \sqrt{17} \approx 4.12$ unit.

Distance between A and B changed by $10.05 - 4.12 = 5.93$ unit .
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