Rectangle J′K′L′M′ shown on the grid is the image of rectangle JKLM after transformation. The same transformation will be applie
d on trapezoid STUV. Rectangle JKLM is drawn on the grid with vertices J at negative 7, negative 6. K is at negative 4, negative 6. L is at negative 4, negative 2. M is at negative 7, negative 2. Rectangle J prime K prime L prime M prime is drawn with vertices J at prime 6, negative 9. K prime is at 9, negative 9. L prime 9, negative 5. M prime is at 6, negative 5. Trapezoid STUV is drawn with vertices at S 3, 2. T is at 5, 5. U is at 2, 6. V is at 1, 4.
What will be the location of T′ in the image trapezoid S′T′U′V′?
The transformed figure has the opposite orientation (A'B'C'D' is counterclockwise, whereas ABCD is clockwise), so a reflection is involved. The segment C'D' is still at the bottom, as is CD, so the reflection must be across a vertical line.
In the attached, the light-blue figure A''B''C''D'' shows the reflection of ABCD across the y-axis. It is easy to see that figure A'B'C'D' is a translation of that 5 units downward.
The appropriate choice is the 2nd one ...
... Reflection over the y-axis followed by a translation down by 5 units. droiyhjfgmdp[ytijkmrdo[ijy[hjmpoiujhokmonj-oimog[fujyjh[opifj[yuj'hkm[ofidyjhoijpofyijh[foist[Yaj[hkmcf[oiyjnjc[city[huff[pay[Pj[ifcjty[ijh[fjotijyh[Hj[fogyish[coif[t0yju[pick[fujph[fj0yh-hifcj[oijhpoijcoijhpouihughdxoiuhtoutpgoindxhouthughxdiurht09uhdxpjrtnhiuhdxiourth8iuthojucfptui9yhufchouihgcifuygtidhiougocdiudrhtudxhfogiurhtiohgcdoiuhoiuxdrhtoiughoiutdgr8uhseoiuhtdiuhroiuiudxdhgioudhugtoirugiudghxtiugdifurtyioudfxhriuytoduyhxotuyotiugoiutyoiuhrtuoihesoitughedugyhoiduhrtouhoiguhoidruthuihgdoiuhrtouighdoiurhtghoidutrhourthiuhdcoiutrioduhtgoiuhoitdrhgiouhoiuhderytgiudyhi8yetgiydhrugterudsoseoysrtgoivdxuht8ygfodushty8ugf9d8sythuygsd9e8tgoyfsdgtofidugsyt8g9s8ewgutiyges8tgiueysg7iygterysgt8t6yoesut
