to compare the
triangles, first we will determine the distances of each side
<span>Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
</span>Solving
<span>∆ABC A(11, 6),
B(5, 6), and C(5, 17)</span>
<span>AB = 6 units BC = 11 units AC = 12.53 units
</span><span>
∆XYZ X(-10, 5), Y(-12, -2), and Z(-4, 15)</span><span>XY = 7.14 units YZ = 18.79 units XZ = 11.66 units</span>
<span>∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).</span>
<span>MN = 6 units NO = 11 units MO = 12.53 units
</span><span>
∆JKL J(17, -2), K(12, -2), and L(12, 7).</span><span>JK = 5 units KL = 9 units JL = 10.30 units
</span><span>
∆PQR P(12, 3), Q(12, -2), and R(3, -2)</span><span>PQ = 5 units QR = 9 units PR = 10.30 units</span>
Therefore
<span>we have the <span>∆ABC and the </span><span>∆MNO </span><span>
with all three sides equal</span> ---------> are congruent
</span><span>we have the <span>∆JKL </span>and the <span>∆PQR
</span>with all three sides equal ---------> are congruent </span>
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
<span> then </span>∆MNO<span> ≅</span> <span>∆ABC </span>
a) Reflexion (x axis)
The coordinate notation for the Reflexion is
(x,y)---- >(x,-y)
<span>∆MNO
M(-9, -4), N(-3, -4), and O(-3, -15).</span>
<span>M(-9, -4)-----------------> M1(-9,4)</span>
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)
<span>∆M1N1O1
M1(-9, 4), N1(-3, 4), and O1(-3, 15).</span>
<span>M1(-9, -4)-----------------> M2(9,4)</span>
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)
<span>∆M2N2O2
M2(9,4), N2(3,4), and O2(3, 15).</span>
<span>M2(9, 4)-----------------> M3(11,6)=A</span>
N2(3,4)------------------ > N3(5,6)=B
O2(3,15)----------------- > O3(5,17)=C
<span>∆ABC A(11, 6),
B(5, 6), and C(5, 17)</span>
∆MNO reflection------- > ∆M1N1O1 reflection---- >
∆M2N2O2
translation -- --> ∆M3N3O3
The ∆M3N3O3=∆ABC
<span>
Therefore ∆MNO ≅ <span>
∆ABC - > </span>
check list</span>
2) If ∆JKL -- by a sequence of rotation and translation--- It
can be obtained ----->∆PQR
<span> then </span>∆JKL ≅
<span>∆PQR </span>
d) Rotation 90 degree anticlockwise
The coordinate notation for the Rotation is
(x,y)---- >(-y, x)
<span>∆JKL J(17, -2), K(12, -2), and L(12, 7).</span>
<span>J(17, -2)-----------------> J1(2,17)</span>
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)
<span>∆J1K1L1 J1(2, 17), K1(2, 12), and L1(-7, 12).</span>
<span>J1(2, 17)-----------------> J2(12,3)=P</span>
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
<span>
Therefore ∆JKL ≅ <span>
∆PQR - > </span><span>
check list</span></span>
. Or we could lose the negative on the 3 in the denominator like this: 
and 
. There you go! Any one of those last three would work. It just depends upon how your teacher wants you to express your answer.