Justify Whether Two Figures Are Congruent | Grade 8 Math | LevelKite

You already know how to describe a sequence of transformations that moves one figure onto another. In this lesson you will justify whether two figures are congruent. That means you decide yes or no, then back up your answer with exact evidence.

Translations, reflections and rotations are rigid motions. A translation slides a figure. A reflection flips it across a mirror line. A rotation turns it about a fixed point. These motions can change where a figure is or which way it faces, but they keep every length and every angle measure the same. The figure a motion produces is its image. Two figures are congruent when some sequence of rigid motions moves one figure exactly onto the other. Each vertex and the vertex it lands on are corresponding vertices. This lesson also calls them partners.

Every figure in this lesson sits on a coordinate plane. The horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis. Lengths are in coordinate units. One unit is the distance from one whole number to the next along an axis.

A horizontal segment has endpoints with the same $y$ -coordinate, so its length is the distance between their $x$ -coordinates. A vertical segment has endpoints with the same $x$ -coordinate, so its length is the distance between their $y$ -coordinates. To find that distance, subtract and take the absolute value. A distance is never negative. For example, from $(4, 1)$ to $(1, 1)$ the length is $|1 - 4| = |-3| = 3$ units.

All coordinates here are exact whole numbers, so every length you find is exact, not an estimate. Base each decision on these numbers, not on how a figure looks. A picture can suggest an answer, but only exact coordinates and the lengths found from them can justify it.

Key concept

Justifying congruence

Two figures are congruent exactly when some sequence of rigid motions moves one onto the other. Rigid motions keep every length and angle measure, so you can justify either answer with exact coordinates.

Yes: name a sequence and check that every vertex lands on its corresponding vertex. For a polygon that is enough. A rigid motion carries each side to the segment between the images of its endpoints, so the sides match too.
No: show a length or angle measure that would have to match but does not.

Example 1: triangle ABC has

A(1, 1)

,

B(4, 1)

and

C(1, 3)

; triangle DEF has

D(6, 3)

,

E(9, 3)

and

F(6, 5)

.

Worked example

Example 1: Triangles that face the same way

Triangle ABC has vertices $A(1, 1)$ , $B(4, 1)$ and $C(1, 3)$ . Triangle DEF has vertices $D(6, 3)$ , $E(9, 3)$ and $F(6, 5)$ . Are the triangles congruent? Justify your answer.

Measure the sides you can read exactly. AB and DE are horizontal, so subtract x-coordinates. AC and DF are vertical, so subtract y-coordinates. The matching lengths suggest pairing A with D, B with E and C with F. Matching lengths are only a clue. The motion will be the proof.
$AB = |4 - 1| = 3,\quad DE = |9 - 6| = 3,\quad AC = |3 - 1| = 2,\quad DF = |5 - 3| = 2$
Check which way each triangle faces. B is 3 units right of A and C is 2 units above A. E is 3 units right of D and F is 2 units above D. The triangles face the same way, so try a translation. A translation slides every point the same distance in the same direction. To move A onto D, subtract the coordinates of A from those of D.
$(6 - 1, 3 - 1) = (5, 2)$
A translation by (5, 2) moves each point (x, y) to (x + 5, y + 2), which is 5 units right and 2 units up. Apply it to every vertex, not only A, and compare each image with its partner.
$A(1, 1) \to (1 + 5, 1 + 2) = (6, 3) = D,\quad B(4, 1) \to (4 + 5, 1 + 2) = (9, 3) = E,\quad C(1, 3) \to (1 + 5, 3 + 2) = (6, 5) = F$
All three vertices land on their partners, so every side lands too. This matches what a translation does. It adds the same 5 to both x-coordinates of A and B, so their difference, which is the length of AB, does not change.
$(4 + 5) - (1 + 5) = 9 - 6 = 3 = 4 - 1$

Answer

Yes. The translation by $(5, 2)$ moves $A$ , $B$ and $C$ exactly onto $D$ , $E$ and $F$ , so triangle ABC is congruent to triangle DEF.

Example 2: triangle ABC has

A(-6, 1)

,

B(-2, 1)

,

C(-6, 4)

; triangle DEF has

D(7, 4)

,

E(3, 4)

,

F(7, 7)

. Reflecting ABC across the

y

-axis gives

A'(6, 1)

,

B'(2, 1)

,

C'(6, 4)

, not drawn.

Worked example

Example 2: Triangles that face different ways

Triangle ABC has vertices $A(-6, 1)$ , $B(-2, 1)$ and $C(-6, 4)$ . Triangle DEF has vertices $D(7, 4)$ , $E(3, 4)$ and $F(7, 7)$ . Are the triangles congruent? Justify your answer.

Measure the horizontal sides AB and DE and the vertical sides AC and DF. The matching lengths suggest pairing A with D, B with E and C with F.
$AB = |-2 - (-6)| = |-2 + 6| = 4,\quad DE = |3 - 7| = |-4| = 4,\quad AC = |4 - 1| = 3,\quad DF = |7 - 4| = 3$
Use signs to check direction. A positive change in x means right, and a negative one means left. B is 4 units right of A, but E is 4 units left of D. Up and down already agree, since C is 3 units above A and F is 3 units above D. A translation keeps every segment pointing the same way, so no translation alone can work.
$\text{A to B: } -2 - (-6) = 4,\qquad \text{D to E: } 3 - 7 = -4$
Reflect across the y-axis, the vertical line x = 0. This reflection moves each point (x, y) to (-x, y). Each point keeps its height and its distance from the y-axis, and a point off the axis moves to the other side. This reverses left and right but keeps up and down, which is exactly the change needed. Apply it to every vertex and call the image A′B′C′.
$A(-6, 1) \to (-(-6), 1) = A'(6, 1),\quad B(-2, 1) \to (-(-2), 1) = B'(2, 1),\quad C(-6, 4) \to (-(-6), 4) = C'(6, 4)$
Check that the reflection keeps lengths even though it changes signs. The x-coordinates of A and B both switch sign, so their difference switches from 4 to -4. A distance is the absolute value of the difference, so A′B′ is still 4 units. The y-coordinates do not change, so A′C′ is still 3 units.
$A'B' = |2 - 6| = |-4| = 4,\quad A'C' = |4 - 1| = 3$
Now B′ is 4 units left of A′, just as E is 4 units left of D, so a translation can finish the job. Moving A′ onto D takes (7 - 6, 4 - 1) = (1, 3). A translation by (1, 3) moves each point (x, y) to (x + 1, y + 3). Apply it to every vertex of A′B′C′.
$A'(6, 1) \to (6 + 1, 1 + 3) = (7, 4) = D,\quad B'(2, 1) \to (2 + 1, 1 + 3) = (3, 4) = E,\quad C'(6, 4) \to (6 + 1, 4 + 3) = (7, 7) = F$

Answer

Yes. Reflecting across the $y$ -axis and then translating by $(1, 3)$ moves $A$ , $B$ and $C$ exactly onto $D$ , $E$ and $F$ , so triangle ABC is congruent to triangle DEF.

Any sequence that lands every vertex on its partner justifies a yes, even if it differs from the one in Example 2. A no works differently. If one sequence you try misses, the figures might still be congruent, because another sequence could work. A side whose length matches no side of the other figure does prove a no. No rigid motion changes a length, so that side has nowhere to land.

Example 3: rectangle ABCD has

A(-7, 1)

,

B(-1, 1)

,

C(-1, 3)

,

D(-7, 3)

; rectangle EFGH has

E(2, 1)

,

F(6, 1)

,

G(6, 4)

,

H(2, 4)

. Points D and H are the corners without labels.

Worked example

Example 3: Two rectangles

Rectangle ABCD has vertices $A(-7, 1)$ , $B(-1, 1)$ , $C(-1, 3)$ and $D(-7, 3)$ . Rectangle EFGH has vertices $E(2, 1)$ , $F(6, 1)$ , $G(6, 4)$ and $H(2, 4)$ . Are the rectangles congruent? Justify your answer.

Every side of ABCD is horizontal or vertical, so find each length by subtracting coordinates and taking the absolute value. ABCD is 6 units long and 2 units tall.
$AB = |-1 - (-7)| = |-1 + 7| = 6,\quad BC = |3 - 1| = 2,\quad CD = |-7 - (-1)| = |-7 + 1| = |-6| = 6,\quad DA = |1 - 3| = |-2| = 2$
Do the same for EFGH. It is 4 units long and 3 units tall.
$EF = |6 - 2| = 4,\quad FG = |4 - 1| = 3,\quad GH = |2 - 6| = |-4| = 4,\quad HE = |1 - 4| = |-3| = 3$
Suppose some sequence of rigid motions did move ABCD onto EFGH. Rigid motions keep every length, so the longest sides of the two rectangles would have to correspond. They do not match. AB and CD are 6 units, but EF and GH are only 4 units. No side of EFGH is 6 units long, so side AB has nowhere to land, whatever sequence you try.
$AB = 6 \neq 4 = EF$

Answer

No. Rectangle ABCD has sides $6$ units long, but no side of rectangle EFGH is longer than $4$ units, so no sequence of rigid motions can move one onto the other. The rectangles are not congruent.

To justify congruence, let exact coordinates do the work. For a yes, name a sequence of rigid motions and check that every vertex lands on its corresponding vertex. For a no, find a length or angle measure that cannot match. A similar look, an equal area or one failed attempt is not proof on its own.