Question

What is the image point of left bracket, minus, 1, comma, 1, right bracket(−1,1) after a translation right 2 units and down 1 unit?

Answer 1

What is the image point of left bracket, 5, comma, 2, right bracket(5,2) after a translation left 1 unit and down 2 units?

Answer 2

What is the image point of left bracket, 6, comma, 8, right bracket(6,8) after a translation right 5 units and up 3 units?

Answer 3

What is the image point of left bracket, 0, comma, 4, right bracket(0,4) after a translation right 2 units and down 3 units?

Answer 4

What is the image of left bracket, 1, comma, 6, right bracket(1,6) after a reflection over the y-axis?

Answer 5

What is the image of left bracket, minus, 1, comma, 5, right bracket(−1,5) after a reflection over the line y, equals, xy=x?

Answer 6

What is the image of (−1,5) after a reflection over the line y=x?

Answer 7

What is the image of (−8,−5) after a reflection over the line y=x?

Answer 8

What is the image of (−8,6) after a reflection over the line y=−x?

Answer 9

What is the image of the point (−9,−1) after a rotation of 270 ∘
counterclockwise about the origin?

Answer 10

What is the image of the point (−1,−4) after a rotation of 90 degrees
counterclockwise about the origin?

Answer 11

What is the image point of left bracket, minus, 3, comma, minus, 8, right bracket(−3,−8) after the transformation T, start subscript, 3, comma, 5, end subscript, circle, R, start subscript, 180, degrees, end subscriptT
3,5
​
∘R
180
∘

​
?

Answer 12

What is the image of the point left bracket, 5, comma, minus, 9, right bracket(5,−9) after a rotation of 180, degrees180
∘
counterclockwise about the origin?

Answer 13

What is the image of the point left bracket, 3, comma, 4, right bracket(3,4) after a rotation of 90, degrees90
∘
counterclockwise about the origin?

Answer 14

What is the image of the point left bracket, minus, 4, comma, 8, right bracket(−4,8) after a rotation of 90, degrees90
∘
counterclockwise about the origin?

Answer 15

What is the image of the point left bracket, minus, 7, comma, minus, 8, right bracket(−7,−8) after a rotation of 180, degrees180
∘
counterclockwise about the origin?

Answer 16

What is the image of the point left bracket, 5, comma, minus, 9, right bracket(5,−9) after a rotation of 270, degrees270
∘
counterclockwise about the origin?

Answer 17

What is the image of the point left bracket, 0, comma, 2, right bracket(0,2) after a rotation of 90, degrees90
∘
counterclockwise about the origin?

Answer 18

What is the image of the point left bracket, 1, comma, 8, right bracket(1,8) after a rotation of 180, degrees180
∘
counterclockwise about the origin?

Answer 19

What is the image of the point left bracket, 0, comma, 5, right bracket(0,5) after a rotation of 180, degrees180
∘
counterclockwise about the origin?

Answer 20

What is the image of the point (0,5) after a rotation of 180 degrees
counterclockwise about the origin?

Answer 21

The image point of (-1,1) after a translation right 2 units and down 1 unit would be (1,0).

Answer 22

The image point of (5,2) after a translation left 1 unit and down 2 units would be (4,0).

Answer 23

The image point of (6,8) after a translation right 5 units and up 3 units would be (11, 11).

Answer 24

The image point of (0,4) after a translation right 2 units and down 3 units would be (2,1).

Answer 25

The image of (1,6) after a reflection over the y-axis would be (-1, 6).

Answer 26

To find the image of a point after a reflection over a line given by the equation y = f(x), we can find the equation of the perpendicular line to f(x) passing through the point. Then we find the intersection point of the two lines, and the image point will be the reflection of the original point over this intersection point.

The equation xy = x can be rearranged as y = 1/x.

The slope of y = 1/x is -1 (by taking the derivative of y = 1/x), so the slope of the perpendicular line is 1.

The equation of the perpendicular line passing through (-1, 5) is y - 5 = 1(x - (-1)), which simplifies to y = x + 6.

To find the intersection point of y = 1/x and y = x + 6, we can substitute y = x + 6 into the equation y = 1/x:

x + 6 = 1/x
(x + 6)x = 1
x^2 + 6x - 1 = 0

Solving this quadratic equation will give us the x-coordinate of the intersection point. Using the quadratic formula, we have:

x = (-6 ± √(6^2 - 4(1)(-1))) / 2
x = (-6 ± √(36 + 4)) / 2
x = (-6 ± √40) / 2
x = (-6 ± 2√10) / 2
x = -3 ± √10

Taking the positive root, x = -3 + √10.

Substituting this back into y = 1/x, we have:

y = 1/(-3 + √10)
y = 1/(-3 + √10) * (-3 - √10)/(-3 - √10)
y = (-3 - √10)/(-9 + 10)
y = (-3 - √10)/1
y = -3 - √10

So the intersection point is (-3 + √10, -3 - √10).

The image point of (-1, 5) after a reflection over the line y = xy = x is the reflection of (-1, 5) over (-3 + √10, -3 - √10).

To find this reflection point, we can calculate the coordinates of the midpoint between (-1,5) and (-3 + √10, -3 - √10):

x-coordinate of midpoint = (x1 + x2) / 2 = (-1 + (-3 + √10)) / 2 = -2 + √10/2
y-coordinate of midpoint = (y1 + y2) / 2 = (5 + (-3 - √10)) / 2 = 1 - √10/2

So the reflection point is (-2 + √10/2, 1 - √10/2).

Finally, the image point of (-1, 5) after the reflection is the reflection of (-1, 5) over (-2 + √10/2, 1 - √10/2), which is:

(2((-2 + √10)/2) - (-1), 2(1 - √10/2) - 5)
Simplifying, we get:

((√10 - 2), -√10 - 1)

Answer 27

To find the image of a point after a reflection over the line y = x, we simply swap the x-coordinate with the y-coordinate of the original point.

Therefore, the image of (-1, 5) after a reflection over the line y = x is (5, -1).

Answer 28

The image of (-8, -5) after a reflection over the line y = x is (-5, -8).

Answer 29

To find the image of a point after a reflection over the line y = -x, we swap the x-coordinate with the negation of the y-coordinate.

So, the image of (-8, 6) after a reflection over the line y = -x is (6, -8).

Answer 30

To find the image of a point after a rotation of 270 degrees counterclockwise about the origin, we can use the following rotation matrix:

[ cos(θ) -sin(θ) ]
[ sin(θ) cos(θ) ]

In this case, θ = 270 degrees.

Applying the rotation matrix to the point (-9, -1):

[ cos(270°) -sin(270°) ] [ -9 ]
[ sin(270°) cos(270°) ] [ -1 ]

The cosine and sine of 270 degrees are 0 and -1 respectively:

[ 0 -(-1) ] [ -9 ]
[ -1 0 ] [ -1 ]

Simplifying:

[ 0 1 ] [ -9 ] = [ 1 ]
[ -1 0 ] [ -1 ] [ -1 ]

Therefore, the image of the point (-9, -1) after a rotation of 270 degrees counterclockwise about the origin is (1, -1).

Answer 31

To find the image of a point after a rotation of 90 degrees counterclockwise about the origin, we can use the following rotation matrix:

[ cos(θ) -sin(θ) ]
[ sin(θ) cos(θ) ]

In this case, θ = 90 degrees.

Applying the rotation matrix to the point (-1, -4):

[ cos(90°) -sin(90°) ] [ -1 ]
[ sin(90°) cos(90°) ] [ -4 ]

The cosine and sine of 90 degrees are 0 and 1 respectively:

[ 0 -(1) ] [ -1 ]
[ 1 0 ] [ -4 ]

Simplifying:

[ 0 -1 ] [ -1 ] = [ 1 ]
[ 1 0 ] [ -4 ] [ -4 ]

Therefore, the image of the point (-1, -4) after a rotation of 90 degrees counterclockwise about the origin is (1, -4).

Answer 32

The transformation T₃,₅∘R₁₈₀∘ applied to the point (-3, -8) involves two steps:

1. Translation by the vector (3, 5):
(-3, -8) + (3, 5) = (0, -3)

2. Rotation by 180 degrees about the origin:
The rotation of (0, -3) by 180 degrees results in (-0, 3), which simplifies to (0, 3).

Therefore, the image point of (-3, -8) after the transformation T₃,₅∘R₁₈₀∘ is (0, 3).

Answer 33

To find the image of a point after a rotation of 180 degrees counterclockwise about the origin, we can use the following rotation matrix:

[ cos(θ) -sin(θ) ]
[ sin(θ) cos(θ) ]

In this case, θ = 180 degrees.

Applying the rotation matrix to the point (5, -9):

[ cos(180°) -sin(180°) ] [ 5 ]
[ sin(180°) cos(180°) ] [ -9 ]

The cosine and sine of 180 degrees are -1 and 0 respectively:

[ -1 -(0) ] [ 5 ]
[ 0 -1 ] [ -9 ]

Simplifying:

[ -1 0 ] [ 5 ] = [ -5 ]
[ 0 -1 ] [ -9 ] [ 9 ]

Therefore, the image of the point (5, -9) after a rotation of 180 degrees counterclockwise about the origin is (-5, 9).

Answer 34

To find the image of a point after a rotation of 90 degrees counterclockwise about the origin, we can use the following rotation matrix:

[ cos(θ) -sin(θ) ]
[ sin(θ) cos(θ) ]

In this case, θ = 90 degrees.

Applying the rotation matrix to the point (3, 4):

[ cos(90°) -sin(90°) ] [ 3 ]
[ sin(90°) cos(90°) ] [ 4 ]

The cosine and sine of 90 degrees are 0 and 1 respectively:

[ 0 -(1) ] [ 3 ]
[ 1 0 ] [ 4 ]

Simplifying:

[ 0 -1 ] [ 3 ] = [ -4 ]
[ 1 0 ] [ 4 ] [ 3 ]

Therefore, the image of the point (3, 4) after a rotation of 90 degrees counterclockwise about the origin is (-4, 3).

Answer 35

To find the image of a point after a rotation of 90 degrees counterclockwise about the origin, we can use the following rotation matrix:

[ cos(θ) -sin(θ) ]
[ sin(θ) cos(θ) ]

In this case, θ = 90 degrees.

Applying the rotation matrix to the point (-4, 8):

[ cos(90°) -sin(90°) ] [ -4 ]
[ sin(90°) cos(90°) ] [ 8 ]

The cosine and sine of 90 degrees are 0 and 1 respectively:

[ 0 -(1) ] [ -4 ]
[ 1 0 ] [ 8 ]

Simplifying:

[ 0 -1 ] [ -4 ] = [ 8 ]
[ 1 0 ] [ 8 ] [ -4 ]

Therefore, the image of the point (-4, 8) after a rotation of 90 degrees counterclockwise about the origin is (8, -4).

Answer 36

To find the image of a point after a rotation of 180 degrees counterclockwise about the origin, we can use the following rotation matrix:

[ cos(θ) -sin(θ) ]
[ sin(θ) cos(θ) ]

In this case, θ = 180 degrees.

Applying the rotation matrix to the point (-7, -8):

[ cos(180°) -sin(180°) ] [ -7 ]
[ sin(180°) cos(180°) ] [ -8 ]

The cosine and sine of 180 degrees are -1 and 0 respectively:

[ -1 -(0) ] [ -7 ]
[ 0 -1 ] [ -8 ]

Simplifying:

[ -1 0 ] [ -7 ] = [ 7 ]
[ 0 -1 ] [ -8 ] [ 8 ]

Therefore, the image of the point (-7, -8) after a rotation of 180 degrees counterclockwise about the origin is (7, 8).

Answer 37

To find the image of a point after a rotation of 270 degrees counterclockwise about the origin, we can use the following rotation matrix:

[ cos(θ) -sin(θ) ]
[ sin(θ) cos(θ) ]

In this case, θ = 270 degrees.

Applying the rotation matrix to the point (5, -9):

[ cos(270°) -sin(270°) ] [ 5 ]
[ sin(270°) cos(270°) ] [ -9 ]

The cosine and sine of 270 degrees are 0 and -1 respectively:

[ 0 -(1) ] [ 5 ]
[ -1 0 ] [ -9 ]

Simplifying:

[ 0 -1 ] [ 5 ] = [ -9 ]
[ -1 0 ] [ -9 ] [ -5 ]

Therefore, the image of the point (5, -9) after a rotation of 270 degrees counterclockwise about the origin is (-9, -5).

Answer 38

To find the image of a point after a rotation of 90 degrees counterclockwise about the origin, we can use the following rotation matrix:

[ cos(θ) -sin(θ) ]
[ sin(θ) cos(θ) ]

In this case, θ = 90 degrees.

Applying the rotation matrix to the point (0, 2):

[ cos(90°) -sin(90°) ] [ 0 ]
[ sin(90°) cos(90°) ] [ 2 ]

The cosine and sine of 90 degrees are 0 and 1 respectively:

[ 0 -(1) ] [ 0 ]
[ 1 0 ] [ 2 ]

Simplifying:

[ 0 -1 ] [ 0 ] = [ -1 ]
[ 1 0 ] [ 2 ] [ 0 ]

Therefore, the image of the point (0, 2) after a rotation of 90 degrees counterclockwise about the origin is (-1, 0).

Answer 39

To find the image of a point after a rotation of 180 degrees counterclockwise about the origin, we can use the following rotation matrix:

[ cos(θ) -sin(θ) ]
[ sin(θ) cos(θ) ]

In this case, θ = 180 degrees.

Applying the rotation matrix to the point (1, 8):

[ cos(180°) -sin(180°) ] [ 1 ]
[ sin(180°) cos(180°) ] [ 8 ]

The cosine and sine of 180 degrees are -1 and 0 respectively:

[ -1 -(0) ] [ 1 ]
[ 0 -1 ] [ 8 ]

Simplifying:

[ -1 0 ] [ 1 ]