To translate the triangle 3 units up, we simply add 3 to the y-coordinates of each vertex:
S: (1, 1) → (1, 4)
T: (2, -3) → (2, 0)
U: (4, 0) → (4, 3)
Therefore, the vertices of the image triangle are S' (1, 4), T' (2, 0), and U' (4, 3).
A triangle has vertices at s ( 1,1) ,T(2,-3),and U(4,0). The triangle is translated up to three units what are the coordinates of the diversities of the image
9 answers
A(5, 3), B(2, 1), and C(-2, 4) are the coordinates of a triangle's vertices. If the triangle is translated right 5 units, what are the coordinates of the image?
A'(5, 8), B'(2, 6), C'(-2, 9)
A'(0, 3), B'(-3, 1), C'(-7, 4)
A'(10, 3), B'(7, 1), C'(3, 4)
A'(10, 8), B'(7, 6), C'(3, 9)
A'(5, 8), B'(2, 6), C'(-2, 9)
A'(0, 3), B'(-3, 1), C'(-7, 4)
A'(10, 3), B'(7, 1), C'(3, 4)
A'(10, 8), B'(7, 6), C'(3, 9)
To translate the triangle 5 units to the right, we simply add 5 to the x-coordinates of each vertex:
A: (5, 3) → (10, 3)
B: (2, 1) → (7, 1)
C: (-2, 4) → (3, 4)
Therefore, the vertices of the image triangle are A' (10, 3), B' (7, 1), and C' (3, 4).
So, the correct answer is A'(10, 3), B'(7, 1), C'(3, 4).
A: (5, 3) → (10, 3)
B: (2, 1) → (7, 1)
C: (-2, 4) → (3, 4)
Therefore, the vertices of the image triangle are A' (10, 3), B' (7, 1), and C' (3, 4).
So, the correct answer is A'(10, 3), B'(7, 1), C'(3, 4).
R(3, 2), S(5, -2), and T(6, 0) are the coordinates of a triangle's vertices. If the triangle is translated down 6 units, what are the coordinates of the image?
R'(3, 8), S'(5, 4), T'(6, 6)
R'(-3, -4), S'(-1, -8), T'(0, -6)
R'(-3, 2), S'(-1, -2), T'(0, 0)
R'(3, -4), S'(5, -8), T'(6, -6)
R'(3, 8), S'(5, 4), T'(6, 6)
R'(-3, -4), S'(-1, -8), T'(0, -6)
R'(-3, 2), S'(-1, -2), T'(0, 0)
R'(3, -4), S'(5, -8), T'(6, -6)
To translate the triangle 6 units down, we simply subtract 6 from the y-coordinates of each vertex:
R: (3, 2) → (3, -4)
S: (5, -2) → (5, -8)
T: (6, 0) → (6, -6)
Therefore, the vertices of the image triangle are R' (3, -4), S' (5, -8), and T' (6, -6).
So, the correct answer is R'(3, -4), S'(5, -8), T'(6, -6).
R: (3, 2) → (3, -4)
S: (5, -2) → (5, -8)
T: (6, 0) → (6, -6)
Therefore, the vertices of the image triangle are R' (3, -4), S' (5, -8), and T' (6, -6).
So, the correct answer is R'(3, -4), S'(5, -8), T'(6, -6).
Point T' has coordinates (-3, 4). If it was translated up 3 units, what were the coordinates of its pre-image?
(-6, 4)
(0, 4)
(-3, 1)
(-3, 7)
(-6, 4)
(0, 4)
(-3, 1)
(-3, 7)
To translate T' up 3 units, we simply subtract 3 from its y-coordinate:
T': (-3, 4) → (-3, 7)
Therefore, the coordinates of T's pre-image must be (-3, 7) since this is the point that was translated to get T'.
So, the correct answer is (-3, 7).
T': (-3, 4) → (-3, 7)
Therefore, the coordinates of T's pre-image must be (-3, 7) since this is the point that was translated to get T'.
So, the correct answer is (-3, 7).
A figure is translated down 6 units. How will the coordinates of the vertices of the image be different from the coordinates of the vertices of the pre-image?
The y-coordinates will decrease by 6.
The x-coordinates will increase by 6.
The x-coordinates will decrease by 6.
The y-coordinates will increase by 6.
The y-coordinates will decrease by 6.
The x-coordinates will increase by 6.
The x-coordinates will decrease by 6.
The y-coordinates will increase by 6.
The correct answer is: The y-coordinates will decrease by 6.
Translation involves moving a figure a certain distance in a certain direction. In this case, the direction of the translation is down, which means that the figure is moving in the negative y-axis direction.
To achieve this, we simply subtract 6 from the y-coordinates of each vertex of the pre-image. So, the y-coordinates of the vertices of the image will be 6 units less than the y-coordinates of the vertices of the pre-image.
Translation involves moving a figure a certain distance in a certain direction. In this case, the direction of the translation is down, which means that the figure is moving in the negative y-axis direction.
To achieve this, we simply subtract 6 from the y-coordinates of each vertex of the pre-image. So, the y-coordinates of the vertices of the image will be 6 units less than the y-coordinates of the vertices of the pre-image.