Question

Quadrilateral ABCD with vertices A(0,7) B(1,3), C(-1,-4), and D(-5,1): <7,-3>

Answer 1

We will have the following:

2)

A(0, 7) : <7, -3>

`A^(\prime)(7,4)`

B(1, 3) : <7, -3>

`B^(\prime)(8,0)`

C(-1, -4) : <7, -3>

`C^(\prime)(6,-7)`

D(-5, 1) : <7, -3>

`D^(\prime)(2,-2)`

3)

From the graph we will have the following:

a.

`(x,y)\to(x+7,y+5)`

b.

`\langle7,5\rangle`

***Explanation***

For point 2, we will simply apply the vector to the corresponding coordinates, that is:

We have the coordinates:

`A(a,b)`

and the vector:

`\langle c,d\rangle`

So, in order to determine the final image we will have to follow the transformation rule:

`A^(\prime)(a+c,b+d)`

*For point 3, we will simply count the number of units the image has moved to the left or rigth and that will be our transformation rule for the x-axis, and the number of units the image has moved up or down and that will be our transformation rule for the y-axis.

In the case of the problem, the images moved 7 units to the rigth (+7) and then moved 5 units up (+5), so the transformation rule in coordinate notation is given by:

`(x,y)\to(x+7,y+5)`

And in order to write it in vector notation, we simply write the units the images move:

`\langle7,5\rangle`