QUESTION IMAGE
Question
write the new coordinates
write the coordinates obtained after the given reflection.
- e(-7,5), f(-3,6), g(-3,2), h(-7,1)
reflection across the line y = - 1
e: __, f: __
g: __, h: __
- c(5,-4), d(8,-5), e(9,-8), f(6,-7)
reflection across the line x = 5
c: __, d: __
e: __, f: __
- a(5,-1), b(3,-4), c(3,0)
reflection across the y - axis
a: __, b: __
c: ____
- p(1,2), q(1,3), r(4,3), s(4,2)
reflection across the line y = 2
p: __, q: __
r: __, s: __
- j(-5,3), k(-1,1), l(-2,3), m(-1,4)
reflection across the line y = x
j: __, k: __
l: __, m: __
- t(5,4), u(6,2), v(8,1), w(7,3)
reflection across the line x = - 1
t: __, u: __
v: __, w: __
- r(-1,-4), s(0,-3), t(-1,-2), u(-2,-3)
reflection across the line x = - 2
r: __, s: __
t: __, u: __
- l(1,1), m(0,-1), n(5,1)
reflection across the line y = 3
l: __, m: __
n: ____
Step1: Recall reflection rules
For reflection across $y = k$: $(x,y)\to(x,2k - y)$. For reflection across $x = h$: $(x,y)\to(2h - x,y)$. For reflection across $y=x$: $(x,y)\to(y,x)$. For reflection across $y - axis$: $(x,y)\to(-x,y)$.
Step2: Solve 1)
Given points $E(-7,5),F(-3,6),G(-3,2),H(-7,1)$ and line $y=-1$.
For point $E$: $x=-7,y = 5,k=-1$, new $y=2\times(-1)-5=-7$, so $E'(-7,-7)$.
For point $F$: $x=-3,y = 6,k=-1$, new $y=2\times(-1)-6=-8$, so $F'(-3,-8)$.
For point $G$: $x=-3,y = 2,k=-1$, new $y=2\times(-1)-2=-4$, so $G'(-3,-4)$.
For point $H$: $x=-7,y = 1,k=-1$, new $y=2\times(-1)-1=-3$, so $H'(-7,-3)$.
Step3: Solve 2)
Given points $C(5,-4),D(8,-5),E(9,-8),F(6,-7)$ and line $x = 5$.
For point $C$: $x = 5,y=-4,h = 5$, new $x=2\times5 - 5=5$, so $C'(5,-4)$.
For point $D$: $x = 8,y=-5,h = 5$, new $x=2\times5 - 8 = 2$, so $D'(2,-5)$.
For point $E$: $x = 9,y=-8,h = 5$, new $x=2\times5 - 9=1$, so $E'(1,-8)$.
For point $F$: $x = 6,y=-7,h = 5$, new $x=2\times5 - 6 = 4$, so $F'(4,-7)$.
Step4: Solve 3)
Given points $A(5,-1),B(3,-4),C(3,0)$ and reflection across $y - axis$.
For point $A$: $x = 5,y=-1$, new $x=-5$, so $A'(-5,-1)$.
For point $B$: $x = 3,y=-4$, new $x=-3$, so $B'(-3,-4)$.
For point $C$: $x = 3,y=0$, new $x=-3$, so $C'(-3,0)$.
Step5: Solve 4)
Given points $P(1,2),Q(1,3),R(4,3),S(4,2)$ and line $y = 2$.
For point $P$: $x = 1,y=2,k = 2$, new $y=2\times2 - 2=2$, so $P'(1,2)$.
For point $Q$: $x = 1,y=3,k = 2$, new $y=2\times2 - 3=1$, so $Q'(1,1)$.
For point $R$: $x = 4,y=3,k = 2$, new $y=2\times2 - 3=1$, so $R'(4,1)$.
For point $S$: $x = 4,y=2,k = 2$, new $y=2\times2 - 2=2$, so $S'(4,2)$.
Step6: Solve 5)
Given points $J(-5,3),K(-1,1),L(-2,3),M(-1,4)$ and line $y=x$.
For point $J$: $(x,y)\to(y,x)$, so $J'(3,-5)$.
For point $K$: $(x,y)\to(y,x)$, so $K'(1,-1)$.
For point $L$: $(x,y)\to(y,x)$, so $L'(3,-2)$.
For point $M$: $(x,y)\to(y,x)$, so $M'(4,-1)$.
Step7: Solve 6)
Given points $T(5,4),U(6,2),V(8,1),W(7,3)$ and line $x=-1$.
For point $T$: $x = 5,y=4,h=-1$, new $x=2\times(-1)-5=-7$, so $T'(-7,4)$.
For point $U$: $x = 6,y=2,h=-1$, new $x=2\times(-1)-6=-8$, so $U'(-8,2)$.
For point $V$: $x = 8,y=1,h=-1$, new $x=2\times(-1)-8=-10$, so $V'(-10,1)$.
For point $W$: $x = 7,y=3,h=-1$, new $x=2\times(-1)-7=-9$, so $W'(-9,3)$.
Step8: Solve 7)
Given points $R(-1,-4),S(0,-3),T(-1,-2),U(-2,-3)$ and line $x=-2$.
For point $R$: $x=-1,y=-4,h=-2$, new $x=2\times(-2)-(-1)=-3$, so $R'(-3,-4)$.
For point $S$: $x = 0,y=-3,h=-2$, new $x=2\times(-2)-0=-4$, so $S'(-4,-3)$.
For point $T$: $x=-1,y=-2,h=-2$, new $x=2\times(-2)-(-1)=-3$, so $T'(-3,-2)$.
For point $U$: $x=-2,y=-3,h=-2$, new $x=2\times(-2)-(-2)=-2$, so $U'(-2,-3)$.
Step9: Solve 8)
Given points $L(1,1),M(0,-1),N(5,1)$ and line $y = 3$.
For point $L$: $x = 1,y=1,k = 3$, new $y=2\times3 - 1=5$, so $L'(1,5)$.
For point $M$: $x = 0,y=-1,k = 3$, new $y=2\times3-(-1)=7$, so $M'(0,7)$.
For point $N$: $x = 5,y=1,k = 3$, new $y=2\times3 - 1=5$, so $N'(5,5)$.
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