Motion. Parallel Translation

Let $\vec{v} = (a,\, b)$ be a fixed vector. The parallel translation by $\vec{v}$ is the transformation that maps each point $A(x, y)$ to the point

$\boxed{A'(x + a,\; y + b)}$

We write $T_{\vec{v}}(A) = A'$, or equivalently $\overrightarrow{AA'} = \vec{v}$ for every point $A$.

Let $T_{\vec{v}}$ be the translation by $\vec{v} = (a, b)$. Then:

1. Isometry: $|A'B'| = |AB|$ for all points $A$, $B$, so $T_{\vec{v}}$ is a motion.

2. Direct isometry: $T_{\vec{v}}$ preserves orientation.

3. Lines map to parallel lines: If $\ell$ is any line, its image $\ell' = T_{\vec{v}}(\ell)$ is parallel to $\ell$ (or equal to $\ell$ when $\vec{v} \parallel \ell$).

4. Every figure maps to a congruent figure: The image of any triangle, polygon, or circle under $T_{\vec{v}}$ is congruent to the original.

5. Composition: Applying $T_{\vec{u}}$ followed by $T_{\vec{v}}$ gives the translation $T_{\vec{u}+\vec{v}}$: $T_{\vec{v}} \circ T_{\vec{u}} = T_{\vec{u}+\vec{v}}$

Proof of (1). Let $A(x_1, y_1)$ and $B(x_2, y_2)$. Then $A' = (x_1+a,\, y_1+b)$ and $B' = (x_2+a,\, y_2+b)$.

$|A'B'| = \sqrt{(x_2+a - x_1-a)^2 + (y_2+b - y_1-b)^2} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = |AB|. \quad \blacktriangleleft$

Triangle $ABC$ has vertices $A(1, 2)$, $B(4, 2)$, $C(3, 5)$. Find the vertices of its image under the translation by $\vec{v} = (-2,\, 3)$.

Solution. Apply $(x, y) \mapsto (x - 2,\; y + 3)$ to each vertex:

$A' = (1-2,\; 2+3) = (-1,\; 5)$ $B' = (4-2,\; 2+3) = (2,\; 5)$ $C' = (3-2,\; 5+3) = (1,\; 8)$

Verification: $|AB| = 3$ and $|A'B'| = |2-(-1)| = 3$ ✓; $|BC|= \sqrt{1+9} = \sqrt{10}$ and $|B'C'| = \sqrt{1+9} = \sqrt{10}$ ✓.

Answer: $A'(-1, 5)$, $B'(2, 5)$, $C'(1, 8)$. $\blacktriangleleft$

A translation maps point $P(3, -1)$ to $P'(7, 4)$ and point $Q(0, 2)$ to $Q'(4, 7)$. Find the translation vector and verify it is consistent.

Solution. From the first pair:

$\vec{v} = \overrightarrow{PP'} = (7-3,\; 4-(-1)) = (4,\; 5).$

Check with the second pair: $\overrightarrow{QQ'} = (4-0,\; 7-2) = (4,\; 5)$ ✓.

Answer: The translation vector is $\vec{v} = (4,\, 5)$. $\blacktriangleleft$

A square has vertices $A(0,0)$, $B(3,0)$, $C(3,3)$, $D(0,3)$. Find the vertices of its image under the translation by $\vec{v} = (1,\, -4)$. Then find the translation vector that maps the image back to the original square.

A translation maps $M(-2, 5)$ to $M'(1, 1)$. Find the image of the point $N(6, -3)$ under the same translation. Also find the preimage of the point $K(0, 0)$ under this translation (i.e., the point that maps to $K$).