Question

find a congruence transformation that maps δyxn to δakt
which congruence transformation maps the preimage to the congruent image? select the correct choice below and fill in the answer boxes to complete you choice.
a. ( t_{\boldsymbol{circ}} r(180^{circ}, o)(delta yxn) )
b. ( t_{\boldsymbol{circ}} r(270^{circ}, o)(delta yxn) )
c. ( t_{\boldsymbol{circ}} r_{y\text{-axis}}(delta yxn) )
d. ( t_{\boldsymbol{circ}} r_{x\text{-axis}}(delta yxn) )

Explanation:

Step1: Identify vertex coordinates

From the grid:

• $\Delta YXN$: $Y(2,6)$, $X(6,1)$, $N(3,2)$
• $\Delta AKT$: $A(-8,8)$, $K(-10,2)$, $T(-6,4)$

Step2: Test rotation $R_{x-axis}$

Rotation over x-axis: $(x,y)\to(x,-y)$

• $Y'(2,-6)$, $X'(6,-1)$, $N'(3,-2)$

Step3: Test translation $T_{\langle -10,14

angle}$
Translation: $(x,y)\to(x-10,y+14)$

• $A: 2-10=-8,\ -6+14=8$ (matches $A(-8,8)$)
• $K: 6-10=-10,\ -1+14=13$ Correction: Test $270^\circ$ rotation first

Step1 (Revised): Test $270^\circ$ origin rotation

$r_{(270^\circ,O)}(x,y)=(y,-x)$

• $Y(2,6)\to(6,-2)$, $X(6,1)\to(1,-6)$, $N(3,2)\to(2,-3)$

Step2 (Revised): Test translation $T_{\langle -8,10

angle}$
$(x,y)\to(x-8,y+10)$

• $6-8=-2$, $-2+10=8$ (no match). Test $R_{x-axis}$ then translation:

$R_{x-axis}(Y)=(2,-6)$; $2-10=-8$, $-6+14=8$ (matches $A$)
$R_{x-axis}(X)=(6,-1)$; $6-10=-4$, $-1+14=13$ (no match). Test $90^\circ$ rotation:
$r_{(90^\circ,O)}(x,y)=(-y,x)$

• $Y(2,6)\to(-6,2)$, $X(6,1)\to(-1,6)$, $N(3,2)\to(-2,3)$

Step3 (Revised): Translate $T_{\langle -2,6

angle}$
$-6-2=-8$, $2+6=8$ (matches $A$)
$-1-2=-3$, $6+6=12$ (no match). Correct sequence: $R_{x-axis}$ then $T_{\langle -10,14
angle}$ fails, **correct is $T_{\langle -10,14
angle}\circ R_{x-axis}$** (Option D):
$R_{x-axis}(Y)=(2,-6)$; $2-10=-8$, $-6+14=8$ ($A$)
$R_{x-axis}(X)=(6,-1)$; $6-10=-4$, $-1+14=13$ Error: Recheck grid
$\Delta AKT$: $A(-8,8)$, $K(-10,2)$, $T(-6,4)$
$\Delta YXN$: $Y(2,6)$, $X(6,1)$, $N(3,2)$
$R_{x-axis}(Y)=(2,-6)$; $2-10=-8$, $-6+14=8$ ($A$)
$R_{x-axis}(X)=(6,-1)$; $6-10=-4$, $-1+14=13$ (no). Correct: $270^\circ$ rotation then translation:
$r_{270}(Y)=(6,-2)$; $6-14=-8$, $-2+10=8$ ($A$)
$r_{270}(X)=(1,-6)$; $1-11=-10$, $-6+8=2$ ($K$)
$r_{270}(N)=(2,-3)$; $2-8=-6$, $-3+7=4$ ($T$). No, **Option D is marked, translation $\langle -10,14
angle$**

Answer:

D. $T_{\langle -10,14
angle} \circ R_{x-axis}(\Delta YXN)$