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Pre-image point N(6, -3) was dilated to point N'(2, -1). What was the scale factor used?
What is the midpoint between (-8, 5) and (2, -2)?

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kudzordzifrancis

Question 1

Let the scale factor be k.

Then, we have the mapping

[tex]N(6,-3)\to N'(6k,-3k)[/tex]

This implies that:

[tex](6k,-3k)=(2,-1)[/tex]

We equate any corresponding component find the value of the scale factor k.

6k=2

[tex]k=\frac{2}{6}[/tex]

[tex]k=\frac{1}{3}[/tex]

Hence the scale factor is  [tex]\frac{1}{3}[/tex]

Question 2:

The midpoint of any two points can be calculated using the formula;

[tex](\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})[/tex]

We want to find the midpoint of (-8, 5) and (2, -2).

[tex](\frac{-8+2}{2}, \frac{5+-2}{2})[/tex]

[tex](\frac{-6}{2}, \frac{3}{2})[/tex]

The midpoint is:

[tex](-3, \frac{3}{2})[/tex]

absor201

Answer:

Scale Factor = (-4,2)

Mid-point = (-3, 3/2)

Step-by-step explanation:

Let X = (x,y) be the scale factor by which the point was dilated. Using this assumption we can write that:

N+X=N^'

(6,-3)+(x,y)=(2,-1)

(6+x,-3+y)=(2,-1)

Putting the coordinates equal one by one:

For x-coordinate:

6+x=2

x=2-6

x= -4

For y-coordinate:

-3+y= -1

y= -1+3

y=2

So the scale factor was:

(-4,2)

For the mid-point of (-8,5) and (2,-2)

Mid-point=((-8+2)/2,(5-2)/2)

=((-6)/2,3/2)

=(-3,3/2)