Question

Rectangle LMNO has vertices L(–4,6), M(–1,6), N(–1,2), and O(–4,2). Suppose you first reflect this rectangle across the y-axis. Then, translate it down four units and to the left one unit. Where are the corresponding vertices L′M′N′O′ located?L′(–5,–10), M′(0,–10), N′(–2,–2), O′(–5,–6)L′(3,2), M′(0,2), N′(0,–2), O′(3,–2)L′(–5,–10), M′(–2,–10), N′(–2,–6), O′(–5,–2)L′(3,2), M′(0,–2), N′(0,2), O′(3,–2)

Answer 1

L′(3,2), M′(0,2), N′(0,–2), O′(3,–2)

Explanation

Step 1

plot the rectangle

Step 2

make the indicated transformations

a)

first reflect this rectangle across the y-axis:

The rule for the reflection around the Y axis is you basically just put a "negative" in front of the X coordinates of each point

so

Let

$\begin{gathered} L(-4,6) \\ M(-1,6) \\ N(-1,2) \\ O(-4,2) \end{gathered}$

then

$\begin{gathered} L(-4,6)\Rightarrow L(-(-4),6)\Rightarrow L1=(4,6) \\ M(-1,6)\Rightarrow M(-(-1),6)\Rightarrow M1=(1,6) \\ N(-1,2)\Rightarrow N(-(-1),2)\Rightarrow N1=(1,2) \\ O(-4,2)\Rightarrow O(-(-4),2)\Rightarrow O1=(4,2) \end{gathered}$

b) Then, translate it down four units and to the left one unit

To translate the point P(x,y) , a units right and b units up, use P'(x+a,y+b)

, if the traslation is down a is negative, if the translation is to the left, b is negative

hence

i) subtract 4 to the y-coordinate

ii)subtract 1 unit from x -axis ,so

$(x,y)\Rightarrow(x-1,y-4)$

therefore,

$\begin{gathered} L1(4,6)\Rightarrow L^{\prime}=(4-1,6-4))=L^{\prime}(3,2) \\ M1(1,6)\Rightarrow M^{^{\prime}}=(1-1,6-4))=L^{\prime}(0,2) \\ N1(1,2)\Rightarrow N^{\prime}=(1-1,2-4))=N^{\prime}(0,-2) \\ O1(4,2)\Rightarrow O(4-1,2-4)\Rightarrow O1=O^{\prime}(3,-2) \end{gathered}$

so,

the answer is

L′(3,2), M′(0,2), N′(0,–2), O′(3,–2)

I hope this helps you