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

Quadrilateral ABCD with vertices A(0,7) B(1,3), C(-1,-4), and D(-5,1): <7,-3&gt-example-1
User Zon
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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

User Atlantis
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