# Point A is at (-8 ,-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?

Jul 8, 2017

The new coordinates are $= \left(- 7 , 8\right)$ and the distance has changed by $= 6.82$

#### Explanation:

The rotation of $\frac{\pi}{2}$ clockwise about the origin transforms the point $A$ into $A '$

The coordinates of $A '$ are

$\left(\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right) \cdot \left(\begin{matrix}- 8 \\ - 7\end{matrix}\right) = \left(\begin{matrix}- 7 \\ 8\end{matrix}\right)$

Distance $A B$ is

$= \sqrt{{\left(- 3 + 8\right)}^{2} + {\left(- 4 + 7\right)}^{2}}$

$= \sqrt{34}$

Distance $A ' B$ is

$= \sqrt{{\left(- 3 + 7\right)}^{2} + {\left(- 4 - 8\right)}^{2}}$

$= \sqrt{160}$

The distance has changed by

$= \sqrt{160} - \sqrt{34}$

$= 6.82$

Jul 8, 2017

New coordinates of point A is (-8,7) and distance between points A and B has changed.

#### Explanation:

When point A is rotated Π/2 clockwise, it moves up and crosses X axis.

For, the original angle created by point A at origin with X axis was less than Π/2.

Now, the distance OA (where, O is origin) is a constant.

Let us designate A' as new position of point A, after rotating Π/2 clockwise.

So we get an isosceles triangle AA'O, where angle AOA' is Π/2.

An isosceles triangle having Π/2 as the angle at vertex, will cause base angles as Π/4.

Now base of the triangle AA'O cuts X axis. Let us designate this point as O'

In triangles A'O'O and AO'O, OO' is a common arm, ∠O'A'O = ∠O'AO =Π/4; for triangle A'OA is an isosceles triangle.

So, triangle A'O'O and triangle AO'O are equal in all respects.

Hence O'A'=O'A=7 units; for coordinate of A was (-8, -7).

So coordinate of A' will be (-8, +7).

Now, distance between two points in Cartesian Coordinates is
${\left({\left(X 1 - X 2\right)}^{2} + {\left(Y 1 - Y 2\right)}^{2}\right)}^{\frac{1}{2}}$, where X1, Y1 and X2,Y2 are the coordinates of the two points.

Original distance was ((-7+3)^2+(-8+4)^2)^.5 =${32}^{.5}$

New distance is ((-7+3)^2+(+8+4)^2)^.5 =${160}^{.5}$

Therefore, we can say that distance between points A and B has changed.