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Instructions:Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17). ∆XYZ has vertices at X(-10, 5), Y(-12, -2), and Z(-4, 15). ∆MNO has vertices at M(-9, -4), N(-3, -4), and O(-3, -15). ∆JKL has vertices at J(17, -2), K(12, -2), and L(12, 7). ∆PQR has vertices at P(12, 3), Q(12, -2), and R(3, -2). can be shown to be congruent by a sequence of reflections and translations. can be shown to be congruent by a single rotation.

Respuesta :

calculista

to compare the triangles, first we will determine the distances of each side

Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
Solving 

∆ABC  A(11, 6), B(5, 6), and C(5, 17)

AB = 6 units   BC = 11 units AC = 12.53 units
∆XYZ  X(-10, 5), Y(-12, -2), and Z(-4, 15)
XY = 7.14 units   YZ = 18.79 units XZ = 11.66 units

∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).

MN = 6 units   NO = 11 units MO = 12.53 units
∆JKL  J(17, -2), K(12, -2), and L(12, 7).
JK = 5 units   KL = 9 units JL = 10.30 units
∆PQR  P(12, 3), Q(12, -2), and R(3, -2)
PQ = 5 units   QR = 9 units PR = 10.30 units 
Therefore
we have the ∆ABC   and the ∆MNO   
with all three sides equal ---------> are congruent  
we have the ∆JKL  and the ∆PQR 
with all three sides equal ---------> are congruent  

 let's check

 Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

 1)     If ∆MNO   ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC 

 then ∆MNO ≅ ∆ABC   

 a)      Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).

M(-9, -4)----------------->  M1(-9,4)

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

 b)      Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

∆M1N1O1  M1(-9, 4), N1(-3, 4), and O1(-3, 15).

M1(-9, -4)----------------->  M2(9,4)

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

 c)   Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

∆M2N2O2  M2(9,4), N2(3,4), and O2(3, 15).

M2(9, 4)----------------->  M3(11,6)=A

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

∆ABC  A(11, 6), B(5, 6), and C(5, 17)

 ∆MNO  reflection------- >  ∆M1N1O1  reflection---- > ∆M2N2O2  translation -- --> ∆M3N3O3 

 The ∆M3N3O3=∆ABC 

Therefore ∆MNO ≅ ∆ABC   - > check list

 2)     If ∆JKL  -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR 

 then ∆JKL ≅ ∆PQR   

 d)     Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

∆JKL  J(17, -2), K(12, -2), and L(12, 7).

J(17, -2)----------------->  J1(2,17)

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

 e)      translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

∆J1K1L1  J1(2, 17), K1(2, 12), and L1(-7, 12).

J1(2, 17)----------------->  J2(12,3)=P

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

 ∆PQR  P(12, 3), Q(12, -2), and R(3, -2)

 ∆JKL  rotation------- >  ∆J1K1L1  translation -- --> ∆J2K2L2=∆PQR 

Therefore ∆JKL ≅ ∆PQR   - > check list
jason357536

Answer:

ABC-MNO and JKL-PQR

Step-by-step explanation:

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