Point A is at #(8 ,-4 )# and point B is at #(2 ,-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

#A(8,-4)to(4,8)," change"≈5.097#

Under a clockwise rotation about the origin of #(3pi)/2#

a point (x ,y) → (-y ,x)

Hence A (8 ,-4) → A(4 ,8)

To calculate the distance (d) between A and B use the #color(blue)"distance formula"#
#color(red)(bar(ul(|color(white)(2/2)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)color(white)(2/2)|)))# where # (x_1,y_1)" and " (x_2,y_2)" are 2 coordinate points"#

The 2 points here are A(8 ,-4) and B(2 ,-3). That is the original A and B.

let # (x_1,y_1)=(8,-4)" and " (x_2,y_2)=(2,-3)#
#d=sqrt((2-8)^2+(-3+4)^2)=sqrt37≈6.083#

Now use the 'new' point A(4 ,8) and B(2 ,-3)

#d=sqrt((2-4)^2+(-3-8)^2)=sqrt125≈11.180#

change in distance between A and B = 11.180 - 6.083 = 5.097

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Answer 2

To find the new coordinates of point A after rotating it clockwise by ( \frac{3\pi}{2} ) about the origin, we can use the following rotation formula:

[ x' = x \cdot \cos(\theta) - y \cdot \sin(\theta) ] [ y' = x \cdot \sin(\theta) + y \cdot \cos(\theta) ]

Where ( (x', y') ) are the new coordinates after rotation, ( (x, y) ) are the original coordinates of point A, and ( \theta ) is the angle of rotation (in this case, ( \frac{3\pi}{2} ) radians).

Given that point A is at (8, -4), we substitute these values into the rotation formulas:

[ x' = 8 \cdot \cos\left(\frac{3\pi}{2}\right) - (-4) \cdot \sin\left(\frac{3\pi}{2}\right) ] [ y' = 8 \cdot \sin\left(\frac{3\pi}{2}\right) + (-4) \cdot \cos\left(\frac{3\pi}{2}\right) ]

Evaluate the trigonometric functions:

[ \cos\left(\frac{3\pi}{2}\right) = 0 ] [ \sin\left(\frac{3\pi}{2}\right) = -1 ]

Substitute these values:

[ x' = 8 \cdot 0 - (-4) \cdot (-1) = 4 ] [ y' = 8 \cdot (-1) + (-4) \cdot 0 = -8 ]

So, the new coordinates of point A after rotation are (4, -8).

To find the change in distance between points A and B, we first calculate the distances between the original points A and B, and then between the new points A and B. The difference between these distances gives us the change.

The distance formula between two points ( (x_1, y_1) ) and ( (x_2, y_2) ) is:

[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

Given points A and B:

Original coordinates of point A: (8, -4) Coordinates of point B: (2, -3)

[ \text{Distance}_\text{original} = \sqrt{(2 - 8)^2 + (-3 - (-4))^2} = \sqrt{(-6)^2 + (1)^2} = \sqrt{36 + 1} = \sqrt{37} ]

New coordinates of point A after rotation: (4, -8)

[ \text{Distance}_\text{new} = \sqrt{(2 - 4)^2 + (-3 - (-8))^2} = \sqrt{(-2)^2 + (5)^2} = \sqrt{4 + 25} = \sqrt{29} ]

The change in distance is given by:

[ \text{Change in distance} = \text{Distance}\text{new} - \text{Distance}\text{original} = \sqrt{29} - \sqrt{37} ]

So, the new coordinates of point A after rotation are (4, -8), and the change in distance between points A and B is approximately ( \sqrt{29} - \sqrt{37} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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