Question

explain 1b performing translat visit bim.easyaccessmaterials.com, read integrated mathematics 1 lesson 11.1, then read the teacher voice - an original figure is often called a preimage. through a transformation, a figure is changed and a new figure, or image, is produced. the inputs for the transformation are the points on the preimage, while the outputs are the points on the image. note that in naming an image, a convention known as prime notation is utilized. for instance, a preimage would be point t, and the image would be t which is known as \t prime.\ translations a translation maps, or moves every point of a figure the same distance in the same direction. for example, the translation in the figure maps all the points of segment (overline{pq}) along a vector (langle a, b
angle) to the segment points (overline{pq}) so that (pp = qq), and (overline{pp} parallel overline{qq}). this means that the points travel along segments that have the same length and are parallel. example 2 - translating a figure using a vector the vertices of (\triangle lmn) are (l(2, 2)), (m(5, 3)), and (n(9, 1)). translate (\triangle lmn) using the vector (langle -2, 6
angle). solution method 1 first, graph (\triangle lmn). use (langle -2, 6
angle) to move each vertex 2 units left and 6 units up. label the image vertices. draw (\triangle lmn). notice that the vectors drawn from the preimage vertices to image vertices are parallel. visit www.bigideasmathvideos.com to watch the flipped video instruction for the \try this\ problem(s) below. try this video for example 2 - translating a figure using a vector 2) the vertices of (\triangle abc) are (a(0, 3)), (b(2, 4)), and (c(1, 0)). translate (\triangle abc) using the vector (langle -1, -2
angle).

Explanation:

Step1: Apply vector to point A

For point $A(0,3)$, add the vector $\langle -1,-2
angle$:
$A'(0 + (-1), 3 + (-2)) = (-1, 1)$

Step2: Apply vector to point B

For point $B(2,4)$, add the vector $\langle -1,-2
angle$:
$B'(2 + (-1), 4 + (-2)) = (1, 2)$

Step3: Apply vector to point C

For point $C(1,0)$, add the vector $\langle -1,-2
angle$:
$C'(1 + (-1), 0 + (-2)) = (0, -2)$

Answer:

The vertices of the translated triangle $\triangle A'B'C'$ are $A'(-1, 1)$, $B'(1, 2)$, and $C'(0, -2)$.
To graph: Plot $A'(-1,1)$, $B'(1,2)$, $C'(0,-2)$ and connect the points to form the triangle.