QUESTION IMAGE
Question
write the coordinates obtained after the giv
- a(0, - 3), b(0, - 5), c(2, - 5), d(3, - 3)
translate 5 units up and 4 units right
a: 4,2 _, b: _
c: 6,0 _, d: _
- k(5, 7), l(3, 7), m(3, 5), n(6, 4)
reflection across the line y = 2
k: _, l: _
m: _, n: _
- u(-3, 6), v(-8, 1), w(-3, 1)
180° rotation about the origin
u: _, v: _
w: _
- r(2, - 6), l(-2, - 8), m(2, - 10)
translate 4 units left and 6 units up
Step1: Translate point B
For a translation of 5 units up and 4 units right, we add 4 to the x - coordinate and 5 to the y - coordinate of point B(0, - 5). So, \(x = 0+4 = 4\) and \(y=-5 + 5=0\).
Step2: Translate point D
For point D(3, - 3), after translation, \(x = 3+4=7\) and \(y=-3 + 5 = 2\).
Step3: Reflect point K across \(y = 2\)
The distance between \(y = 7\) (y - coordinate of K) and \(y = 2\) is \(d=7 - 2=5\). The new y - coordinate of \(K'\) is \(2-5=-3\), and the x - coordinate remains the same. So \(K'=(5,-3)\).
Step4: Reflect point L across \(y = 2\)
The distance between \(y = 7\) (y - coordinate of L) and \(y = 2\) is \(d = 7 - 2=5\). The new y - coordinate of \(L'\) is \(2-5=-3\), and the x - coordinate remains the same. So \(L'=(3,-3)\).
Step5: Reflect point M across \(y = 2\)
The distance between \(y = 5\) (y - coordinate of M) and \(y = 2\) is \(d=5 - 2 = 3\). The new y - coordinate of \(M'\) is \(2-3=-1\), and the x - coordinate remains the same. So \(M'=(3,-1)\).
Step6: Reflect point N across \(y = 2\)
The distance between \(y = 4\) (y - coordinate of N) and \(y = 2\) is \(d=4 - 2=2\). The new y - coordinate of \(N'\) is \(2-2 = 0\), and the x - coordinate remains the same. So \(N'=(6,0)\).
Step7: Rotate point U 180° about the origin
For a 180° rotation about the origin, the rule is \((x,y)\to(-x,-y)\). For U(-3,6), \(U'=(3,-6)\).
Step8: Rotate point V 180° about the origin
For V(-8,1), \(V'=(8,-1)\).
Step9: Rotate point W 180° about the origin
For W(-3,1), \(W'=(3,-1)\).
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B': (4,0)
D': (7,2)
K': (5,-3)
L': (3,-3)
M': (3,-1)
N': (6,0)
U': (3,-6)
V': (8,-1)
W': (3,-1)