Point A is at #(-2 ,8 )# and point B is at #(-1 ,3 )#. 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?

Answer 1

Sadie Allen

The change in distance is:

#Deltad = d_2 - d_1#

#Deltad = sqrt82 - sqrt26#

The first distance from B to A is:

#d_1 = sqrt((-2 - -1)^2 + (8 - 3)^2)#

#d_1 = sqrt((-1)^2 + (5)^2)#

#d_1 = sqrt(1 + 25)#

#d_1 = sqrt26#

Compute the polar coordinates of point A:

#r = sqrt((-2)^2 + (8)^2)#

#r = sqrt(4 + 64)#

#r = sqrt68#

#theta_1 = tan^-1(8/-2) + pi#

Note: We add #pi#, because the inverse tangent function returns either first quadrant angles or negative rotations. In this case, it will return a negative angle.

To rotate the point we add #(3pi)/2# to the original angle:

#theta_2 = tan^-1(8/-2) + (5pi)/2#

Compute the x and y coordinates for new point A:

#(sqrt68cos(tan^-1(8/-2) + (5pi)/2), sqrt68sin(tan^-1(8/-2) + (5pi)/2))#

#(8, 2)#

Compute the new distance:

#d_2 = sqrt((8 - -1)^2 + (2 - 3)^2)#

#d_2 = sqrt((9)^2 + (-1)^2)#

#d_2 = sqrt82#

The change in distance is:

#Deltad = d_2 - d_1#

#Deltad = sqrt82 - sqrt26#

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Answer 2

Noah Atkinson

The new coordinates of point A after rotating $(3\pi)/2$ clockwise about the origin are $(8, 2)$ .

To find the new coordinates, we use the following rotation formulas:

$x' = x \cos(\theta) - y \sin(\theta)$ $y' = x \sin(\theta) + y \cos(\theta)$

For point A $(-2, 8)$ , substituting $\theta = (3\pi)/2$ , we get:

$x' = -2 \cos\left(\frac{3\pi}{2}\right) - 8 \sin\left(\frac{3\pi}{2}\right) = 8$ $y' = -2 \sin\left(\frac{3\pi}{2}\right) + 8 \cos\left(\frac{3\pi}{2}\right) = 2$

So, the new coordinates of point A are $(8, 2)$ .

The original distance between points A and B can be calculated using the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

For points A $(-2, 8)$ and B $(-1, 3)$ , the original distance is:

$d_{\text{original}} = \sqrt{(-1 - (-2))^2 + (3 - 8)^2} = \sqrt{1^2 + 5^2} = \sqrt{26}$

After the rotation, the new coordinates of point A are $(8, 2)$ . So, the distance between the new point A and B $(-1, 3)$ is:

$d_{\text{new}} = \sqrt{(8 - (-1))^2 + (2 - 3)^2} = \sqrt{9^2 + (-1)^2} = \sqrt{82}$

The change in distance is given by:

$\Delta d = d_{\text{new}} - d_{\text{original}} = \sqrt{82} - \sqrt{26}$

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