Math 8 · 8.G.A.1 · 8.G.A.3

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You slide a shape across the page without turning it. How do you say exactly where every point of it ends up?

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Why this exists

Push a paper triangle across a desk without letting it turn. Every corner moves, every edge moves, and yet nothing about the triangle has changed except where it is. Saying “it moved a bit to the right” is not enough to draw the new position. Two people following that sentence would draw two different triangles.

What is needed is a description that settles where every point goes, and words for talking about the shape before and after the move. It turns out two numbers are enough: how far across and how far up. That pair of numbers is a translation.

Translations are the simplest of the motions that move a figure without changing its size or shape. The others are reflections and rotations, and together they are how geometry says two figures are congruent. A translation is also what makes a wallpaper pattern repeat, and what moves a graph to a new place on the coordinate plane.

This lesson runs over two parts. Part 1 sets up the language (preimage, image, mapping) and the coordinate rule for a translation, and uses it in both directions. Part 2 shows what a translation keeps, how to tell whether a copy really is a translation of a figure, how to undo and combine translations, and how a translation moves a whole line.

Definition

Transformation, preimage and image. A transformation of the plane is a rule that sends every point P of the plane to exactly one point P'. The point you start from is the preimage, and the point it is sent to is its image. We say the transformation maps P to P' and write

P \mapsto P'

read “P maps to P prime”. When the transformation has a name, say T, its image of P is written T(P), so T(P) = P'. The image of a figure is the set of images of all its points, and the image of a vertex keeps the vertex’s letter with a prime: the image of triangle ABC is triangle A'B'C'.

Translation. A translation slides every point of the plane the same distance in the same direction. On a coordinate grid it is fixed by two numbers: a, how far each point moves across (to the right when a is positive, to the left when a is negative), and b, how far each point moves up (up when b is positive, down when b is negative). We write the translation as \langle a, b \rangle. Moving across changes only the first coordinate, and moving up changes only the second, so the translation’s coordinate rule is

(x, y) \mapsto (x + a,\ y + b)

The arrow from any point to its image, \overrightarrow{PP'}, is the same arrow for every point of the plane: the same length and the same direction. That arrow is the vector of the translation, and a and b are its across and up parts, its components. This is why the star is called a translation by a vector.

Figure 1The translation \langle 6, 2 \rangle. Every point of the flag makes the same trip, 6 units right and 2 units up.

In Figure 1 the whole flag moves 6 units right and 2 units up. That is the translation \langle 6, 2 \rangle, with rule (x, y) \mapsto (x + 6,\ y + 2). The two dashed arrows are drawn from different points of the flag, but they are copies of one arrow.

The translation \langle 0, 0 \rangle moves nothing at all. It still counts as a translation, the one that leaves every point where it is.

Intuition

Trace a figure onto tracing paper, then slide the paper across the page without turning it. Every mark on the paper travels together, so every mark makes exactly the same trip. That is the whole idea: a translation is one trip, taken by every point at once.

Because the paper never turns and never flips over, a translated figure faces the way it did before. In Figure 2 all four flags have the same size and shape. Flag 2 is flag S slid 9 units right and 1 unit up: its pole is still upright and its triangle still points right. Flag 1 points left, so it is a mirror image of S, and flag 3 lies on its side, so it has been turned. No amount of sliding produces either of those.

Figure 2Four copies of the same size and shape. Only one of them can be reached from S by sliding.

On a grid, across and up do not interfere with each other. Moving right never changes how high a point is, and moving up never changes how far right it is. That independence is why the rule simply adds a to the first coordinate and b to the second.

Results

Part 1: the coordinate rule

The rule. The translation \langle a, b \rangle sends each point to

(x, y) \mapsto (x + a,\ y + b)

One pair of points settles everything. If a translation sends P(x, y) to P'(x', y'), then its vector is

\langle a, b \rangle = \langle x' - x,\ y' - y \rangle

and from that the image of every other point follows. In particular, the image of the origin is (0 + a, 0 + b) = (a, b), so the origin’s image names the translation.

Running it backwards. The preimage of a point (x', y') is (x' - a,\ y' - b). The translation \langle -a, -b \rangle undoes \langle a, b \rangle.

Part 2: what a translation keeps

Lengths. For any two points, P'Q' = PQ.

Directions. The segment \overline{P'Q'} has the same run and the same rise as \overline{PQ}. It is parallel to \overline{PQ} (or lies on the same line) and points the same way.

Angles. Every angle keeps its measure.

So the image of a figure is congruent to it and faces the same way: never flipped, never turned.

Fixed points. \langle 0, 0 \rangle leaves every point where it is. Every other translation moves every point, and leaves no point fixed.

Two in a row. Doing \langle a, b \rangle and then \langle c, d \rangle is the single translation \langle a + c,\ b + d \rangle.

Why it is true

The rule. Moving a units across changes only the first coordinate, by a; moving b units up changes only the second, by b. So (x, y) lands on (x + a, y + b), and since every point makes the same trip, the same a and b serve for every point. If P(x, y) \mapsto P'(x', y') then x' = x + a and y' = y + b, which gives a = x' - x and b = y' - y.

Lengths and directions. Take two points P(x_1, y_1) and Q(x_2, y_2). Their images are P'(x_1 + a,\ y_1 + b) and Q'(x_2 + a,\ y_2 + b). Subtract to get the run and the rise from P' to Q':

\begin{aligned} (x_2 + a) - (x_1 + a) &= x_2 - x_1 \\ (y_2 + b) - (y_1 + b) &= y_2 - y_1 \end{aligned}

The a and the b cancel. The run and rise from P' to Q' are exactly the run and rise from P to Q. By the distance formula,

P'Q' = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = PQ

Equal run and rise also mean equal slope (when the run is not zero), so the two segments are parallel or on one line; when the run is zero, both are vertical. And because the run and rise have the same signs, \overrightarrow{P'Q'} points the same way as \overrightarrow{PQ}.

Figure 3Both arrows are the same arrow, a across and b up, so PQQ'P' is a parallelogram and P'Q' is a copy of PQ.

Figure 3 shows the same fact as a picture. The arrows \overrightarrow{PP'} and \overrightarrow{QQ'} are both a across and b up, so they are equal and parallel; and \overline{P'Q'} is equal and parallel to \overline{PQ}. A quadrilateral whose opposite sides are equal and parallel is a parallelogram, so PQQ'P' is one, unless all four points happen to lie on one line. That parallelogram is the starting point of the more formal definition of a translation in rigid motions defined precisely.

Angles. A triangle PQR goes to a triangle P'Q'R' whose three sides have the same lengths as before, so the two triangles are congruent by side-side-side and their angles match. Any angle can be made into an angle of a triangle by choosing one point on each of its rays, so every angle keeps its measure.

Fixed points. A point stays put when (x + a, y + b) = (x, y), which needs a = 0 and b = 0. Otherwise no point stays put; each moves the same positive distance \displaystyle \sqrt{a^2 + b^2}.

Two in a row. \langle a, b \rangle sends (x, y) to (x + a, y + b), and then \langle c, d \rangle sends that to

(x + a + c,\ y + b + d)

which is the rule for \langle a + c,\ b + d \rangle. Taking c = -a and d = -b gives \langle 0, 0 \rangle, which is why \langle -a, -b \rangle undoes \langle a, b \rangle.

Worked examples

Examples 1 to 4 go with Part 1, the language and the coordinate rule. Examples 5 to 8 go with Part 2: what a translation keeps, recognizing one, combining them, and moving a line.

Example 1

Translating a triangle

Triangle ABC has vertices A(-4, 1), B(-1, 1) and C(-3, 4). Translate it 6 units right and 3 units down. Write the vector and the coordinate rule, and find the vertices of the image.

Figure 4Example 1. Triangle ABC and its image under \langle 6, -3 \rangle.

Solution

Right is the positive direction across, so a = 6. Down is the negative direction up, so b = -3. The translation is \langle 6, -3 \rangle, and its rule adds 6 to every first coordinate and -3 to every second:

(x, y) \mapsto (x + 6,\ y - 3)

Apply the rule to each vertex:

\begin{aligned} A(-4, 1) &\mapsto A'(2, -2) \\ B(-1, 1) &\mapsto B'(5, -2) \\ C(-3, 4) &\mapsto C'(3, 1) \end{aligned}

Only the three vertices need computing. Every other point of the triangle makes the same trip, so the image of each side is the segment joining the images of its ends.

Check. \overline{AB} is horizontal with length -1 - (-4) = 3, and \overline{A'B'} is horizontal with length 5 - 2 = 3. In Figure 4 every dashed arrow is 6 squares right and 3 down.

Example 2

Finding the translation from one point

A translation sends P(3, -2) to P'(-1, 5). Find its vector and its rule, and the images of the origin and of R(-2, -6).

Solution

Every point makes the same trip, so the trip from P to P' is the whole translation. Subtract the coordinates of the preimage from those of the image:

a = -1 - 3 = -4 \qquad b = 5 - (-2) = 7

The translation is \langle -4, 7 \rangle, 4 units left and 7 units up, with rule (x, y) \mapsto (x - 4,\ y + 7).

The origin goes to (0 - 4,\ 0 + 7) = (-4, 7), and R(-2, -6) goes to (-2 - 4,\ -6 + 7) = (-6, 1).

Check. From -2 to -6 is 4 units left, and from -6 to 1 is 7 units up, the same trip as P made.

Example 3

Running the rule backwards

A translation has rule (x, y) \mapsto (x + 5,\ y - 2), and the image of a point D is D'(1, 3). Find D, and find the translation that undoes this one.

Solution

Call the preimage D(x, y). The rule says its image is (x + 5,\ y - 2), and that image is (1, 3). Two coordinates that are equal give two equations:

x + 5 = 1 \qquad y - 2 = 3

so x = -4 and y = 5. The preimage is D(-4, 5).

Check. (-4 + 5,\ 5 - 2) = (1, 3).

Going backwards meant subtracting 5 and adding 2, which is the translation \langle -5, 2 \rangle, with rule (x, y) \mapsto (x - 5,\ y + 2). Doing \langle 5, -2 \rangle and then \langle -5, 2 \rangle brings every point back to where it started.

Example 4

A point named by the translation itself

The translation \langle a, b \rangle sends the point (a, b) to (6, -10). Find the translation, and the image of the origin.

Solution

The rule adds a to the first coordinate and b to the second, so (a, b) goes to (a + a,\ b + b) = (2a, 2b). Setting this equal to (6, -10):

2a = 6 \qquad 2b = -10

so a = 3 and b = -5. The translation is \langle 3, -5 \rangle.

The origin goes to (0 + 3,\ 0 - 5) = (3, -5). That is the vector itself, and it always will be: under \langle a, b \rangle the origin goes to (a, b). The image of the origin is the quickest way to read off a translation.

Check. (3, -5) \mapsto (3 + 3,\ -5 - 5) = (6, -10).

Example 5

Length and slope survive

Segment \overline{AB} joins A(-2, 1) to B(2, 4). Translate it by \langle 5, -6 \rangle. Compare the length and slope of the image with those of \overline{AB}, and find how far each point moves.

Solution

The images are A'(-2 + 5,\ 1 - 6) = A'(3, -5) and B'(2 + 5,\ 4 - 6) = B'(7, -2).

From A to B the run is 2 - (-2) = 4 and the rise is 4 - 1 = 3. From A' to B' the run is 7 - 3 = 4 and the rise is -2 - (-5) = 3. The same run and rise give the same length and the same slope:

AB = A'B' = \sqrt{4^2 + 3^2} = 5 \qquad \text{slope} = \frac{3}{4}

So \overline{A'B'} is parallel to \overline{AB} and just as long, as Part 2 promised.

Each point moves along the vector \langle 5, -6 \rangle, whose length is

\sqrt{5^2 + 6^2} = \sqrt{61} \approx 7.81

so every point of the segment, and every other point of the plane, moves \displaystyle \sqrt{61} units. The four points A, B, B', A' make a parallelogram, as in Figure 3.

Example 6

Is it a translation?

Triangle JKL has vertices J(-1, 1), K(3, 2) and L(0, 4).

(a) Is triangle U(2, -1), V(6, 0), W(3, 2) the image of JKL under a translation that sends J to U, K to V and L to W?

(b) Triangle X(0, 1), Y(-4, 2), Z(-1, 4) has the same three side lengths as JKL. Is it the image of JKL under any translation at all?

Solution

(a) A translation moves every point by the same arrow, so work out the trip each vertex makes:

\begin{aligned} J \to U &: \langle 2 - (-1),\ -1 - 1 \rangle = \langle 3, -2 \rangle \\ K \to V &: \langle 6 - 3,\ 0 - 2 \rangle = \langle 3, -2 \rangle \\ L \to W &: \langle 3 - 0,\ 2 - 4 \rangle = \langle 3, -2 \rangle \end{aligned}

All three trips are the same, so yes: UVW is the image of JKL under \langle 3, -2 \rangle. All three had to be checked. One pair of points only tells you which translation to try; it is the other vertices that confirm it.

(b) A translation keeps the run and rise of every side, so the image of \overline{JK}, which runs 4 and rises 1, must be a side of the new triangle whose ends differ by exactly \langle 4, 1 \rangle. List the sides of XYZ in both directions:

Side	One way	The other way
\overline{XY}	X \to Y: \langle -4, 1 \rangle	Y \to X: \langle 4, -1 \rangle
\overline{YZ}	Y \to Z: \langle 3, 2 \rangle	Z \to Y: \langle -3, -2 \rangle
\overline{ZX}	Z \to X: \langle 1, -3 \rangle	X \to Z: \langle -1, 3 \rangle

None is \langle 4, 1 \rangle, so no translation carries JKL onto XYZ, whichever vertex is matched to which.

The two triangles are still congruent. The rule (x, y) \mapsto (-x - 1,\ y) sends J, K and L to X, Y and Z: it is the reflection in the y-axis followed by a slide 1 unit left. XYZ is a mirror image of JKL, and a slide cannot produce a mirror image.

Example 7

Two slides and a slide back

Translate by \langle 4, -1 \rangle and then by \langle -7, 5 \rangle. Find the single translation that does the same job, how far it moves each point, and the translation that undoes it.

Solution

Follow a general point through both:

\begin{aligned} (x, y) &\mapsto (x + 4,\ y - 1) \\ &\mapsto (x + 4 - 7,\ y - 1 + 5) \\ &= (x - 3,\ y + 4) \end{aligned}

The combined move is \langle -3, 4 \rangle: the across parts add, 4 + (-7) = -3, and the up parts add, -1 + 5 = 4. Each point moves

\sqrt{(-3)^2 + 4^2} = 5

units. The translation that undoes it is \langle 3, -4 \rangle.

Check with the point (2, 2): the first slide takes it to (6, 1) and the second to (-1, 6), while \langle -3, 4 \rangle takes it straight to (2 - 3,\ 2 + 4) = (-1, 6).

The two separate slides were \displaystyle \sqrt{17} and \displaystyle \sqrt{74} units long, about 4.12 and 8.60, yet the combined trip is only 5. Translations combine by adding their components, not their lengths, because the second trip can partly cancel the first.

Example 8

Translating a line

Translate the line y = 2x - 1 by \langle 3, 5 \rangle and find the equation of its image. Then find every translation that sends this line onto itself.

Solution

A point (X, Y) lies on the image exactly when its preimage lies on the original line. The preimage of (X, Y) is (X - 3,\ Y - 5), so the condition is that these coordinates satisfy y = 2x - 1:

\begin{aligned} Y - 5 &= 2(X - 3) - 1 \\ Y - 5 &= 2X - 7 \\ Y &= 2X - 2 \end{aligned}

The image is the line y = 2x - 2. The slope is still 2, so the image is parallel to the original, as Part 2 says it must be.

Check. (0, -1) is on the original line, and it goes to (3, 4). On the image line, 2(3) - 2 = 4.

Notice the signs. The points moved 3 right and 5 up, yet the equation shows X - 3 and Y - 5, because it describes where the points came from. The same thing happens when a graph is shifted up, down, left or right.

For a general translation \langle a, b \rangle the same reasoning gives

\begin{aligned} Y - b &= 2(X - a) - 1 \\ Y &= 2X - 1 + (b - 2a) \end{aligned}

This is the original line exactly when b - 2a = 0, that is, when b = 2a. Those are the arrows that rise 2 for every 1 across, the arrows that point along the line itself, such as \langle 1, 2 \rangle and \langle -3, -6 \rangle. They slide the line along itself. Every other translation sends it to a different, parallel line.

Boundaries

Every point must make the same trip. The rule (x, y) \mapsto (x + 2,\ 2y) moves the origin by \langle 2, 0 \rangle but moves (0, 3) to (2, 6), a trip of \langle 2, 3 \rangle. Different points make different trips, so it is not a translation, even though it slides some points. A rule is a translation only when it adds the same two numbers to every point.

Slides only. A translated figure faces the same way it did. A congruent copy that is flipped or turned, like flags 1 and 3 in Figure 2, is not the image of the figure under any translation. It can be reached by a reflection or a rotation, possibly combined with a slide, which is the business of composing transformations.

The whole plane moves. A translation is a rule for every point of the plane, not just for the points of the figure. That is why it makes sense to ask for the image of the origin, or of a point nowhere near the figure.

One translation fixes points, the rest fix none. \langle 0, 0 \rangle leaves everything where it is. Any other translation moves every point by the same positive distance.

Parallel, or the same line. A translation sends a line to a parallel line, except when its arrow points along the line; then the line slides along itself and lands on itself (Example 8).

Points and equations move with opposite signs. Points move by +a and +b. An equation for the image replaces x by x - a and y by y - b, because the equation tests where each point came from.

Where this goes next

Translations are the first of the three basic rigid motions. Reflections flip a figure over a line, and rotations turn it about a point; each has its own coordinate rule, and, unlike a translation, each leaves some points fixed. What all three keep, lengths, angles and parallel lines, is collected in what rigid motions preserve, and rigid motions defined precisely builds the translation again from a directed segment and the parallelogram of Figure 3, with no coordinates at all.

Translations combine with the others. Composing transformations follows one motion with another, and two reflections in parallel mirrors turn out to make a translation, as two reflections in a row shows. Sequences of these motions are how congruence is defined: two figures are congruent when some sequence of them carries one onto the other.

The arrow of a translation becomes a vector in its own right, with a length and components, and adding translations becomes adding vectors. On graphs, translating left, right, up and down is the shifting of a graph, and repeating one tile by translations in two directions is the simplest way to build a tessellation.

Exercises

Part 1: images of points

Find the image of P(2, 6) under the translation \langle 3, -4 \rangle.
A translation moves every point 7 units right and 2 units down. Write its vector, and find the image of Q(-6, 0).
A translation has rule (x, y) \mapsto (x - 8,\ y + 3). Find the image of R(5, -5).
Write the vector and the rule of the translation that moves every point 4 units left and 9 units up. What is the image of (4, -9)?
Triangle RST has vertices R(0, 2), S(3, 1) and T(1, 4). Find the vertices of its image under \langle -5, -3 \rangle.
In Figure 1, the tip of the first flag’s triangle is at (-1, 0). Where is the tip of the image’s triangle? Write the rule of the translation.

Part 1: finding the translation

A translation sends (3, 8) to (-2, 10). Find its vector, and the image of (0, -4).
A translation sends the origin to (7, -3). Find the image of (-7, 3).
A translation has rule (x, y) \mapsto (x + 2,\ y - 9). Which point does it send to (4, -1)?
A translation sends G(-1, 4) to G'(5, 4). It sends a point H to H'(0, -6). Find H.
The translation \langle a, b \rangle sends the point (a, 2b) to (10, 9). Find a and b.
A square has vertices (0, 0), (3, 0), (3, 3) and (0, 3). It is translated so that the vertex (3, 3) lands on (-1, 2). Find the vector, and the images of all four vertices.

Part 1: reasoning

Explain why knowing where a translation sends the origin tells you the whole translation.
Hana says that (x, y) \mapsto (x + 3,\ y) and (x, y) \mapsto (x,\ y + 3) are the same translation, because each moves every point 3 units. Is she right? Explain.
Is the rule (x, y) \mapsto (-x,\ y + 5) a translation? Explain, using the points (0, 0) and (2, 0).

Part 2: what a translation keeps

The segment from (-3, -1) to (1, 2) is translated by \langle 4, 5 \rangle. Find the endpoints of the image, and show that the segment and its image have the same length and the same slope.
Points P(1, 1) and Q(4, 5) are translated by \langle 2, -3 \rangle. Find PQ, P', Q' and P'Q', and how far each point moves.
Triangle ABC has vertices A(-3, 1), B(-1, 1) and C(-3, 6). (a) Is the triangle with vertices (1, -2), (3, -2) and (1, 3) the image of ABC under a translation? (b) Is the triangle with vertices (2, 0), (4, 0) and (4, 5)?
P(-4, 2) and Q(6, -8) are translated by \langle 1, 5 \rangle. Find the midpoint M of \overline{PQ} and its image M'. Then find P', Q' and the midpoint of \overline{P'Q'}. What do you notice?

Part 2: undoing and combining

Which translation undoes \langle -6, 11 \rangle?
Translate by \langle 2, 7 \rangle and then by \langle -8, -15 \rangle. Find the single translation that does the same job, how far it moves each point, and the image of (1, 1).
A translation by \langle 5, -1 \rangle is followed by a second translation, and together they send the origin to (-2, 6). Find the second translation.
A translation T sends (2, 3) to (5, 1). Where does doing T three times in a row send (2, 3)?

Part 2: lines and circles

Find the equation of the image of the line y = 3x + 2 under \langle 1, 0 \rangle.
Find the equation of the image of the line y = -x + 4 under \langle 2, 5 \rangle.
Find the translation of the form \langle a, 0 \rangle that sends the line y = 2x + 6 onto the line y = 2x. Then find the translation of the form \langle 0, b \rangle that does the same.
A circle has center (2, -1) and radius 4. Find the center and radius of its image under \langle -5, 3 \rangle.

Part 2: reasoning

Show that a translation keeps midpoints: for any points P and Q, the image of the midpoint of \overline{PQ} is the midpoint of \overline{P'Q'}.
Show that the translation \langle a, b \rangle sends the line y = mx + c onto itself exactly when b = ma.
Show that doing \langle a, b \rangle and then \langle c, d \rangle has the same result as doing \langle c, d \rangle and then \langle a, b \rangle.
Points P and Q are translated to P' and Q'. Explain why PQQ'P' is a parallelogram, and say when it fails to be one.

Answers

P'(5, 2). The rule adds 3 and -4: (2 + 3,\ 6 - 4).
\langle 7, -2 \rangle, and Q'(1, -2). Right 7 gives a = 7, down 2 gives b = -2, and (-6 + 7,\ 0 - 2) = (1, -2).
R'(-3, -2), from (5 - 8,\ -5 + 3).
\langle -4, 9 \rangle, with rule (x, y) \mapsto (x - 4,\ y + 9). The image of (4, -9) is (4 - 4,\ -9 + 9) = (0, 0).
R'(-5, -1), S'(-2, -2) and T'(-4, 1). Subtract 5 from each first coordinate and 3 from each second.
(5, 2), and the rule is (x, y) \mapsto (x + 6,\ y + 2). The flag moves 6 right and 2 up, so the tip goes to (-1 + 6,\ 0 + 2).
\langle -5, 2 \rangle, and (0, -4) \mapsto (-5, -2). The trip is a = -2 - 3 = -5 and b = 10 - 8 = 2.
(0, 0). The origin’s image names the translation, \langle 7, -3 \rangle, and (-7 + 7,\ 3 - 3) = (0, 0).
(2, 8). Solve x + 2 = 4 and y - 9 = -1.
H(-6, -6). The trip from G to G' is \langle 6, 0 \rangle, so go backwards from H': (0 - 6,\ -6 - 0).
a = 5 and b = 3. The image of (a, 2b) is (a + a,\ 2b + b) = (2a, 3b), so 2a = 10 and 3b = 9.
\langle -4, -1 \rangle, from (-1 - 3,\ 2 - 3). The images are (-4, -1), (-1, -1), (-1, 2) and (-4, 2).
Under \langle a, b \rangle the origin goes to (0 + a,\ 0 + b) = (a, b). So its image gives both numbers a and b, and those two numbers fix the rule (x, y) \mapsto (x + a,\ y + b) for every point.
No. Each moves every point 3 units, but in different directions: the first sends (0, 0) to (3, 0) and the second sends it to (0, 3). A translation is a distance and a direction together, and \langle 3, 0 \rangle and \langle 0, 3 \rangle are different arrows.
No. (0, 0) \mapsto (0, 5), a trip of \langle 0, 5 \rangle, but (2, 0) \mapsto (-2, 5), a trip of \langle -4, 5 \rangle. In a translation every point makes the same trip. This rule does keep lengths (those two points are 2 apart, and so are their images), but it flips the plane: it is the reflection in the y-axis followed by a slide 5 units up. Keeping lengths is not enough to make a rule a translation.
The image runs from (1, 4) to (5, 7). Both segments have run 4 and rise 3, so both have length \displaystyle \sqrt{4^2 + 3^2} = 5 and slope \displaystyle \frac{3}{4}.
PQ = 5, P'(3, -2), Q'(6, 2) and P'Q' = 5; each point moves \displaystyle \sqrt{13} \approx 3.61. Both segments have run 3 and rise 4, and the trip \langle 2, -3 \rangle has length \displaystyle \sqrt{2^2 + 3^2}.
(a) Yes, under \langle 4, -3 \rangle: A \mapsto (1, -2), B \mapsto (3, -2) and C \mapsto (1, 3), every vertex making the same trip. (b) No. In ABC the vertex A has one neighbor 2 units right of it and the other 5 units above it, and a translation keeps those trips, so the image needs a vertex with the same two neighbors. From (2, 0) the others are \langle 2, 0 \rangle and \langle 2, 5 \rangle away; from (4, 0) they are \langle -2, 0 \rangle and \langle 0, 5 \rangle; from (4, 5) they are \langle -2, -5 \rangle and \langle 0, -5 \rangle. None works. The triangle is congruent to ABC but is its mirror image.
M(1, -3) and M'(2, 2). P'(-3, 7) and Q'(7, -3), whose midpoint is \displaystyle \left(\frac{-3 + 7}{2},\ \frac{7 - 3}{2}\right) = (2, 2). The image of the midpoint is the midpoint of the image: a translation keeps midpoints.
\langle 6, -11 \rangle, since \langle -6 + 6,\ 11 - 11 \rangle = \langle 0, 0 \rangle.
\langle -6, -8 \rangle, moving each point \displaystyle \sqrt{36 + 64} = 10 units; (1, 1) \mapsto (-5, -7). The components add: 2 - 8 = -6 and 7 - 15 = -8.
\langle -7, 7 \rangle. The first translation takes the origin to (5, -1), and the trip from there to (-2, 6) is \langle -2 - 5,\ 6 + 1 \rangle.
(11, -3). T is \langle 3, -2 \rangle, so three in a row is \langle 9, -6 \rangle, and (2 + 9,\ 3 - 6) = (11, -3).
y = 3x - 1. A point (x, y) is on the image when (x - 1, y) is on the original: y = 3(x - 1) + 2.
y = -x + 11. The preimage of (x, y) is (x - 2,\ y - 5), so y - 5 = -(x - 2) + 4. Check: (0, 4) \mapsto (2, 9), and -2 + 11 = 9.
\langle 3, 0 \rangle and \langle 0, -6 \rangle. Under \langle a, 0 \rangle the image is y = 2(x - a) + 6 = 2x - 2a + 6, which is y = 2x when a = 3. Under \langle 0, b \rangle it is y = 2x + 6 + b, which is y = 2x when b = -6.
Center (-3, 2) and radius 4. The center moves to (2 - 5,\ -1 + 3), and a translation keeps every distance, so every point still lies 4 units from the new center.
Let the translation be \langle a, b \rangle, with P(x_1, y_1) and Q(x_2, y_2). The midpoint M of \overline{PQ} is \displaystyle \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right), so its image is \displaystyle M'\left(\frac{x_1 + x_2}{2} + a,\ \frac{y_1 + y_2}{2} + b\right). The images are P'(x_1 + a,\ y_1 + b) and Q'(x_2 + a,\ y_2 + b), and the midpoint of \overline{P'Q'} has first coordinate \displaystyle \frac{(x_1 + a) + (x_2 + a)}{2} = \frac{x_1 + x_2}{2} + a, and second coordinate \displaystyle \frac{y_1 + y_2}{2} + b the same way. That is M'.
The preimage of (x, y) is (x - a,\ y - b), so the image line is y - b = m(x - a) + c, that is, y = mx + c + (b - ma). This is the original line exactly when b - ma = 0, that is, b = ma: the arrow rises m for every 1 across, so it points along the line.
The first order sends (x, y) to (x + a + c,\ y + b + d) and the second to (x + c + a,\ y + d + b). These are equal because adding numbers in either order gives the same sum, so both orders are the translation \langle a + c,\ b + d \rangle.
\overrightarrow{PP'} and \overrightarrow{QQ'} are the same arrow, so \overline{PP'} and \overline{QQ'} are equal and parallel; and \overline{P'Q'} has the same run and rise as \overline{PQ}, so those two are equal and parallel as well. A quadrilateral with both pairs of opposite sides equal and parallel is a parallelogram. It fails when the arrow points along the line PQ (or is \langle 0, 0 \rangle): then all four points lie on one line and there is no quadrilateral at all.