Question

c) a reflection in the line $y=-1$point $a$ is 4 units above $y=-1$, so $a$ is 4 units below $y=-1$ at $(1,-5)$.similarly, $b$ is 3 units below $y=-1$ at $(5,-4)$, and $c$ is 2 units below $y=-1$ at $(2,-3)$.visit www.bigideasmathvideos.com to watch the flipped video instruction for the \try this\ problem(s) below.try this video for example 1 - reflecting in horizontal and vertical linesgraph $\triangle abc$ with vertices $a(1, 3)$, $b(5, 2)$, and $c(2, 1)$ and its image after the reflection described.1) in the line $n: x=-1$2) in the line $m: y=3$integrated math 1 b credit 4page 12

Explanation:

Step1: Find reflection of A over $x=-1$

Distance from $A(1,3)$ to $x=-1$: $1 - (-1) = 2$. New $x$-coordinate: $-1 - 2 = -3$. $A'(-3, 3)$

Step2: Find reflection of B over $x=-1$

Distance from $B(5,2)$ to $x=-1$: $5 - (-1) = 6$. New $x$-coordinate: $-1 - 6 = -7$. $B'(-7, 2)$

Step3: Find reflection of C over $x=-1$

Distance from $C(2,1)$ to $x=-1$: $2 - (-1) = 3$. New $x$-coordinate: $-1 - 3 = -4$. $C'(-4, 1)$

Step4: Find reflection of A over $y=3$

Distance from $A(1,3)$ to $y=3$: $3 - 3 = 0$. $A'(1, 3)$

Step5: Find reflection of B over $y=3$

Distance from $B(5,2)$ to $y=3$: $3 - 2 = 1$. New $y$-coordinate: $3 + 1 = 4$. $B'(5, 4)$

Step6: Find reflection of C over $y=3$

Distance from $C(2,1)$ to $y=3$: $3 - 1 = 2$. New $y$-coordinate: $3 + 2 = 5$. $C'(2, 5)$

Answer:

1. Reflected vertices over $x=-1$: $A'(-3, 3)$, $B'(-7, 2)$, $C'(-4, 1)$ (plot these and connect to form $\triangle A'B'C'$, along with original $\triangle ABC$)
2. Reflected vertices over $y=3$: $A'(1, 3)$, $B'(5, 4)$, $C'(2, 5)$ (plot these and connect to form $\triangle A'B'C'$, along with original $\triangle ABC$)