Question

use the graph and the translation $(x,y)\to(x + 3,y - 4)$ to answer parts a and b below.
a. enter the image of each vertex as an ordered pair.
$r\to r$ (type an ordered pair.)
$s\to s$ (type an ordered pair.)
$t\to t$ (type an ordered pair.)
$v\to v$ (type an ordered pair.)

Explanation:

Step1: Find the coordinates of original points

Assume the coordinates of \(R\), \(S\), \(T\), \(V\) from the graph. Let's say \(R=(x_1,y_1)\), \(S=(x_2,y_2)\), \(T=(x_3,y_3)\), \(V=(x_4,y_4)\). From the graph, \(R=(2, - 2)\), \(S=(1, - 1)\), \(T=(7, - 1)\), \(V=(6,0)\)

Step2: Apply the translation rule \((x,y)\to(x + 3,y-4)\)

For point \(R\):
\(x\) - coordinate: \(x_1+3=2 + 3=5\)
\(y\) - coordinate: \(y_1-4=-2-4=-6\)
So \(R'\) is \((5,-6)\)

For point \(S\):
\(x\) - coordinate: \(x_2+3=1 + 3=4\)
\(y\) - coordinate: \(y_2-4=-1-4=-5\)
So \(S'\) is \((4,-5)\)

For point \(T\):
\(x\) - coordinate: \(x_3+3=7 + 3=10\)
\(y\) - coordinate: \(y_3-4=-1-4=-5\)
So \(T'\) is \((10,-5)\)

For point \(V\):
\(x\) - coordinate: \(x_4+3=6 + 3=9\)
\(y\) - coordinate: \(y_4-4=0-4=-4\)
So \(V'\) is \((9,-4)\)

Answer:

\(R\to R'(5,-6)\), \(S\to S'(4,-5)\), \(T\to T'(10,-5)\), \(V\to V'(9,-4)\)