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?
39 answers
The image point of (-1,1) after a translation right 2 units and down 1 unit would be (1,0).
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?
The image point of (5,2) after a translation left 1 unit and down 2 units would be (4,0).
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?
The image point of (6,8) after a translation right 5 units and up 3 units would be (11, 11).
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?
The image point of (0,4) after a translation right 2 units and down 3 units would be (2,1).
What is the image of left bracket, 1, comma, 6, right bracket(1,6) after a reflection over the y-axis?
The image of (1,6) after a reflection over the y-axis would be (-1, 6).
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?
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)
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)
What is the image of (−1,5) after a reflection over the line y=x?
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).
Therefore, the image of (-1, 5) after a reflection over the line y = x is (5, -1).
What is the image of (−8,−5) after a reflection over the line y=x?
The image of (-8, -5) after a reflection over the line y = x is (-5, -8).
What is the image of (−8,6) after a reflection over the line y=−x?
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).
So, the image of (-8, 6) after a reflection over the line y = -x is (6, -8).
What is the image of the point (−9,−1) after a rotation of 270 ∘
counterclockwise about the origin?
counterclockwise about the origin?
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).
[ 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).
What is the image of the point (−1,−4) after a rotation of 90 degrees
counterclockwise about the origin?
counterclockwise about the origin?
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).
[ 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).
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
∘
?
3,5
∘R
180
∘
?
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).
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).
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?
∘
counterclockwise about the origin?
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).
[ 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).
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?
∘
counterclockwise about the origin?
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).
[ 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).
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?
∘
counterclockwise about the origin?
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).
[ 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).
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?
∘
counterclockwise about the origin?
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).
[ 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).
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?
∘
counterclockwise about the origin?
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).
[ 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).
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?
∘
counterclockwise about the origin?
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).
[ 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).
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?
∘
counterclockwise about the origin?
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 ]
[ 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 ]
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?
∘
counterclockwise about the origin?
What is the image of the point (0,5) after a rotation of 180 degrees
counterclockwise about the origin?
counterclockwise about the origin?