Translation

The coordinate rule

Translation by vector $(a,b)$ is

For a movement of $5$ units right and $3$ units down, use $(5,-3)$. The point $P(-2,1)$ moves to $P'(3,-2)$ because $-2+5=3$ and $1-3=-2$.

The distance moved is the vector’s magnitude, not the sum of the component magnitudes:

Moving horizontally then vertically would travel $8$ units along that particular two-leg route. The translation vector describes the net displacement, whose straight-line length is $\sqrt{34}$.

Translating a figure

Let a triangle have vertices $A(1,1)$, $B(4,1)$ and $C(2,3)$. Translate it by $(3,-2)$:

Connect the image vertices in corresponding order. The base remains horizontal, the side lengths remain unchanged and the original counterclockwise vertex order remains counterclockwise. You do not need to move the vertices in a special order; every point uses the same rule.

A translation acts on the whole plane, not merely the marked vertices. Because it preserves collinearity, translating endpoints and joining them gives the correct translated edge.

Why distances and slopes stay the same

Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ be arbitrary points. Their images have coordinates $(x_1+a,y_1+b)$ and $(x_2+a,y_2+b)$. Subtracting corresponding coordinates cancels the translation:

The distance formula therefore produces the same distance before and after. For a non-vertical segment, the slope is unchanged for the same reason. Vertical segments stay vertical, even though their slope is undefined.

A non-zero translation has no fixed point, since $x+a=x$ and $y+b=y$ require $a=b=0$. This does not prevent an infinite line from being mapped onto itself by a translation along the line: the set can stay the same while its individual points move.

Finding the vector from a point and its image

If $P(2,-1)$ maps to $P'(7,3)$, subtract original from image:

To claim that an entire labelled figure is a translation, verify that other corresponding pairs give the same vector. If one vertex requires $(5,4)$ and another requires $(4,5)$, no single translation accomplishes both matches.

To recover an original point, subtract the vector. If $Q'(-1,6)$ is the image under $(5,4)$, then $Q=(-1-5,6-4)=(-6,2)$.

Combining and undoing translations

Translation by $(3,-2)$ followed by $(-1,5)$ is equivalent to one translation by

For example, $(1,1)$ first moves to $(4,-1)$ and then to $(3,4)$. Adding the combined vector directly gives the same final point.

The inverse translation uses the opposite vector. Following $(3,-2)$ with $(-3,2)$ gives the zero vector, so every point returns to its original position. Unlike rotation and translation in general, two translations commute: changing their order does not change their vector sum.

Translating a graph is not changing just the output

Suppose a graph is $y=f(x)$ and it is translated by $(a,b)$. A point $(u,f(u))$ becomes $(u+a,f(u)+b)$. Renaming the new horizontal coordinate as $x$ gives $u=x-a$, so the image graph is

For $y=x^2$, a translation of $(3,-2)$ gives $y=(x-3)^2-2$. The minus sign inside the expression comes from recovering the old input; it does not mean the graph moves left.

This connects geometric translation to Transformations of Functions. It is the same point movement described in two different representations.

Interpreting a displacement model

On a plan with one coordinate unit representing $0.5\text{ m}$, vector $(6,-4)$ means $3\text{ m}$ right and $2\text{ m}$ down. The net distance is $\sqrt{13}\text{ m}$, not $\sqrt{52}\text{ m}$; the latter is the numerical distance in coordinate units before conversion.

Keep both axes in the same physical unit when interpreting Euclidean distance. A graph with time on one axis and temperature on the other does not describe a spatial displacement in metres.

Related topics

Vectors in Geometry develops displacement algebra. Compare translation with Rotation and Reflection, which preserve distances but change direction or orientation.