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.

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

∆ABC  A(11, 6), B(5, 6), and C(5, 17)

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

 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 

 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)

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 

i got 10.28 but that rounds up to 10.3