To translate the entire triangle, you first figure out the “move” (translation vector) from the original vertex to its image, then apply that same move to every vertex.

Step-by-step idea

Imagine you have a triangle with vertices
A(x1,y1)A(x_1,y_1)A(x1​,y1​), B(x2,y2)B(x_2,y_2)B(x2​,y2​), and C(x3,y3)C(x_3,y_3)C(x3​,y3​),
and you are told that AAA moves to A′(x1′,y1′)A'(x_1',y_1')A′(x1′​,y1′​) under a translation.

1. Find the translation vector

You ask: “How far right/left and up/down did AAA move to become A′A'A′?”

  • Horizontal move (change in xxx):
    Δx=x1′−x1\Delta x=x_1'-x_1Δx=x1′​−x1​

  • Vertical move (change in yyy):
    Δy=y1′−y1\Delta y=y_1'-y_1Δy=y1′​−y1​

So the translation vector is
(Δx,Δy)=(x1′−x1, y1′−y1)(\Delta x,\Delta y)=(x_1'-x_1,;y_1'-y_1)(Δx,Δy)=(x1′​−x1​,y1′​−y1​). This vector is the rule for the translation.

2. Apply that vector to all vertices

Now you apply this same (Δx,Δy)(\Delta x,\Delta y)(Δx,Δy) to every vertex of the original triangle:

  • A′(x1′,y1′)=(x1+Δx, y1+Δy)A'(x_1',y_1')=(x_1+\Delta x,;y_1+\Delta y)A′(x1′​,y1′​)=(x1​+Δx,y1​+Δy)
    (this matches the given image point)

  • B′(x2′,y2′)=(x2+Δx, y2+Δy)B'(x_2',y_2')=(x_2+\Delta x,;y_2+\Delta y)B′(x2′​,y2′​)=(x2​+Δx,y2​+Δy)

  • C′(x3′,y3′)=(x3+Δx, y3+Δy)C'(x_3',y_3')=(x_3+\Delta x,;y_3+\Delta y)C′(x3′​,y3′​)=(x3​+Δx,y3​+Δy)

That gives you the coordinates of the entire translated triangle.

Quick example story

Think of the translation vector like an instruction: “Move 4 steps right and 2 steps up.” If one vertex of the triangle moves by that rule, every vertex must follow the same rule to keep the shape congruent and in the same orientation. So you don’t need all the image points—just one image vertex is enough to find the “move,” and then you copy that move to the other vertices.

One-sentence answer

Find the translation vector by subtracting the coordinates of the original vertex from its image, then add that same vector to the coordinates of each vertex of the original triangle to get the full translated triangle.