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
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)
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)
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).
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)
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).
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 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.
To find the new coordinates of the vertices after the triangle is translated up by three units, you need to add 3 to the y-coordinates of each vertex.
Let's start by adding 3 to the y-coordinate of vertex S(1,1):
New coordinate of S: S'(1,1+3) = S'(1,4)
Next, add 3 to the y-coordinate of vertex T(2,-3):
New coordinate of T: T'(2,-3+3) = T'(2,0)
Similarly, add 3 to the y-coordinate of vertex U(4,0):
New coordinate of U: U'(4,0+3) = U'(4,3)
Therefore, the coordinates of the vertices after the translation are:
S'(1,4), T'(2,0), U'(4,3).
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).
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).