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

Mar 14, 2018

New coordinate: $\left(- 2 , - 4\right)$
Difference in distance: $\sqrt{5} - 1$

#### Explanation:

Finding the new coordinate
With the $x$ -coordinate positive and $y$ -coordinate negative the point $A$ used to lie in the fourth quadrant of the Cartesian plane (the bottom-right corner); rotating it clockwise by $\frac{\pi}{2}$ (or equivalently by ${90}^{\text{o}}$ ) would move it to the third quadrant (the bottom-left corner) such that both its $x$ and $y$ -coordinates are now positive.

Rotating a point with respect to the origin by an odd integer multiple of $\frac{\pi}{2}$ or ${90}^{\text{o}}$ would swap its $x$ and $y$ -coordinates. So now point $A$ would have coordinates $\left(- 2 , - 4\right)$.

Finding and comparing the distance between $A$ and $B$
Now that we have got the coordinates of $A$ before and after the rotation, so we can apply the Pythagorean theorem in the Cartesian to find the distance between point $A$ and $B$:

Before the rotation:
${A}_{i} B = \sqrt{{\left({x}_{{A}_{i}} - {x}_{B}\right)}^{2} + {\left({y}_{{A}_{i}} - {y}_{B}\right)}^{2}} = \sqrt{{\left(4 - 2\right)}^{2} + {\left(- 2 - \left(- 3\right)\right)}^{2}} = \sqrt{5}$

After the rotation:
${A}_{f} B = \sqrt{{\left({x}_{{A}_{f}} - {x}_{B}\right)}^{2} + {\left({y}_{{A}_{f}} - {y}_{B}\right)}^{2}} = \sqrt{{\left(- 2 - 2\right)}^{2} + {\left(- 4 - \left(- 3\right)\right)}^{2}} = 1$

Therefore the difference would be
$\left\mid {A}_{i} B - {A}_{f} B \right\mid = \sqrt{5} - 1$