# What is the best description of the translation of pre-image abc to image a'b'c'? a'b'c'

###### Question:

a'b'c'

## Answers

the translation is 4 units at right and 2 units up

Step-by-step explanation:

see the attached figure to better understand the problem

looking at the graph

we have

The coordinates of the pre-image ABC are

A(-1,1),B(2,2),C(1,-1)

The coordinates of the image A'B'C' are

A'(3,3),B'(6,4),C'(5,1)

so

A(-1,1) ----> A'(3,3)

A(-1,1) ----> A'(-1+a,1+b)

-1+a=3 ---> a=3+1=4

1+b=3 ---> b=3-1=2

The rule of the translation is

(x,y) ----> (x+4,y+2)

That means ----> the translation is 4 units at right and 2 units up

[tex]What is the best description of the translation of pre-image abc to image a'b'c'? a'b'c'[/tex]

Translation 4 units to the right and 2 units up.

Step-by-step explanation:

[tex]What is the best description of the translation of pre-image abc to image a'b'c'? a'b'c'[/tex]

There is no association.

[tex]Which statement best describes the association between variable x and variable y? perfect negative[/tex]

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

[tex]m_ {1} * m_ {2} = - 1[/tex]

then, given the following line:

[tex]6x-3y = 18\\6x-18 = 3y[/tex]

[tex]\frac {6x} {3} - \frac {18} {3} = y\\y = \frac {6x} {3} - \frac {18} {3}\\y = 2x-6[/tex]

so, [tex]m_ {1} = 2[/tex]

we are looking for m_ {2}:

[tex]m_{2}=\frac{-1}{m_{1}}\\m_{2}=\frac{-1}{2}\\m_{2}=-\frac{1}{2}[/tex]

then, the line is given by:

[tex]y = - \frac {1} {2} x + b[/tex]

we find "b" replacing the given point:

[tex]8 = - \frac {1} {2} (0) + b\\8 = b[/tex]

finally, the equation is:

[tex]y = - \frac {1} {2} x + 8[/tex]

option c

[tex]Me it is a geometry question and you[/tex]