Transformations

In a translation, every point is shifted by the same displacement. That displacement is written as the translation vector:

$\left(\begin{array}{c}a\\ b\end{array}\right)$

So every point moves $a$ units in the horizontal direction and $b$ units in the vertical direction.

The actual distance moved is the straight-line length of the displacement vector. From the IB formula for vector magnitude:

$|v|=\sqrt{v_1^2+v_2^2}$

Applying this to $\left(\begin{array}{c}a\\ b\end{array}\right)$ gives the distance

$d=\sqrt{a^2+b^2}$

Check the other statements:

• Option B: $a-b$ combines components incorrectly; it is not a straight-line distance.
• Option C: $a^2+b^2$ gives $d^2$, not $d$.
• Option D: a translation applies the same displacement everywhere, so the distance cannot depend on the starting point.

Since the displacement length is the magnitude of $\left(\begin{array}{c}a\\ b\end{array}\right)$, the distance every point moves is $\sqrt{a^2+b^2}$.

Take any point $P$ with coordinates $(x,y)$. After a translation by $\left(\begin{array}{c}a\\ b\end{array}\right)$, the $x$-value increases by $a$ and the $y$-value increases by $b$, so the image point is

$P^{\prime }=(x+a,y+b)$

This describes the movement of any point on the object.

Now find the distance $d$ from $P(x,y)$ to $P^{\prime }(x+a,y+b)$ using the standard distance formula:

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Substitute the coordinates:

$d=\sqrt{((x+a)-x{)}^2+((y+b)-y{)}^2}$

Simplify inside the square root:

$\begin{array}{rl}d& =\sqrt{(x+a-x{)}^2+(y+b-y{)}^2}\\ d& =\sqrt{(a{)}^2+(b{)}^2}\\ d& =\sqrt{a^2+b^2}\end{array}$

Because $x$ and $y$ cancel, the distance does not change from point to point.

Using the simplified result:

• Option B is wrong because it does not use the Pythagorean relationship for the diagonal distance.
• Option C is wrong because it gives $d^2$ instead of $d$.
• Option D is wrong because the cancellation shows the distance is independent of $(x,y)$.

The distance moved is the same for every point and equals the magnitude of the translation vector: $\sqrt{a^2+b^2}$.

Choose simple values so we can calculate quickly. Let $a=5$ and $b=12$, giving the translation vector

$\left(\begin{array}{c}5\\ 12\end{array}\right)$

Also choose a starting point such as $(0,0)$.

Add the vector components to the coordinates:

Starting point: $(0,0)$ New point: $(0+5,0+12)=(5,12)$

So the point moves from $(0,0)$ to $(5,12)$.

Use the distance formula:

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Substitute:

$d=\sqrt{(5-0{)}^2+(12-0{)}^2}$ $d=\sqrt{5^2+12^2}$ $d=\sqrt{25+144}=\sqrt{169}=13$

So the distance moved is $13$.

With $a=5$ and $b=12$, compare each claim to $d=13$:

• Option B: $a-b=5-12=-7$, which is not $13$ (and distance can’t be negative).
• Option C: $a^2+b^2=5^2+12^2=169$, which is not $13$.
• Option D: Try another point, like $(1,1)$, which goes to $(6,13)$. The distance $\sqrt{(6-1{)}^2+(13-1{)}^2}$ is still $13$, so it does not depend on the starting coordinates.

Check Option A with the same values:

$\sqrt{a^2+b^2}=\sqrt{5^2+12^2}=\sqrt{169}=13$

This exactly matches the distance we calculated, so the correct statement is that the distance is $\sqrt{a^2+b^2}$.

We are asked to judge what happens to a figure after two transformations in order:

1. A reflection across the $x$-axis.
2. A translation by the vector $\left(\begin{array}{c}2\\ 3\end{array}\right)$.

Our goal is to decide whether the final image stays congruent to the starting pre-image.

An isometry (also called a rigid transformation) keeps all distances and angle measures the same.

So if a transformation is an isometry, the figure’s size and shape are unchanged, meaning the image and pre-image are always congruent. This idea is central in Topic 5: Geometric transformations.

Now we check each transformation:

• Reflection: This flips a figure over a line. Even though the figure may face the opposite way, side lengths and angles do not change, so reflection is an isometry.
• Translation: This slides a figure using a vector. Sliding does not stretch or distort anything, so translation is also an isometry.

Consider two points $A(0,0)$ and $B(1,0)$. Their distance is $1$.

1. After reflecting across the $x$-axis, the points become $A^{\prime }(0,0)$ and $B^{\prime }(1,0)$, and the distance is still $1$.
2. After translating by $\left(\begin{array}{c}2\\ 3\end{array}\right)$, the points become $A^{\prime \prime }(2,3)$ and $B^{\prime \prime }(3,3)$.

We can compute the distance between $A^{\prime \prime }$ and $B^{\prime \prime }$ to check that it remains $1$.

When you perform one isometry after another, the overall transformation is still an isometry. Since both reflection and translation preserve all distances and angles, doing them in sequence also preserves distances and angles everywhere.

So the final image must be congruent to the original pre-image.

Because the transformations used are isometries, the statement is correct, so we select the option that says the image is congruent to the pre-image.

To check whether congruence is preserved, we can test a clear example: a right triangle on the coordinate plane (from Geometry of Shapes: Geometric transformations).

Let the original triangle have vertices:

• $A(0,0)$
• $B(3,0)$
• $C(0,4)$

Its side lengths are $3$, $4$, and $5$.

First we reflect across the $x$-axis, using the rule $(x,y)\to (x,-y)$.

Applying it:

• $A^{\prime }(0,0)$
• $B^{\prime }(3,0)$
• $C^{\prime }(0,-4)$

Next we translate by $\left(\begin{array}{c}2\\ 3\end{array}\right)$, which means $(x,y)\to (x+2,y+3)$.

Applying this to the reflected points:

• $A^{\prime \prime }(0+2,0+3)=(2,3)$
• $B^{\prime \prime }(3+2,0+3)=(5,3)$
• $C^{\prime \prime }(0+2,-4+3)=(2,-1)$

Now compare the side lengths of $A^{\prime \prime }{B}^{\prime \prime }{C}^{\prime \prime }$ to the original ($3,4,5$) using: $\text{distance}=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

• Side $A^{\prime \prime }{B}^{\prime \prime }=\sqrt{(5-2{)}^2+(3-3{)}^2}=\sqrt{3^2+0^2}=3$
• Side $A^{\prime \prime }{C}^{\prime \prime }=\sqrt{(2-2{)}^2+(-1-3{)}^2}=\sqrt{0^2+(-4{)}^2}=4$
• Side $B^{\prime \prime }{C}^{\prime \prime }=\sqrt{(2-5{)}^2+(-1-3{)}^2}=\sqrt{(-3{)}^2+(-4{)}^2}=5$

Because the side lengths match exactly, the final triangle is congruent to the original. Reflection and translation are isometries, so they do not change size or shape.

That means the 'False' option cannot be correct.

This example confirms that reflecting across the $x$-axis and then translating by $\left(\begin{array}{c}2\\ 3\end{array}\right)$ produces an image that remains congruent to the pre-image.