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)?

Question 1

Let the scale factor be k.

Then, we have the mapping

This implies that:

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

6k=2

Hence the scale factor is  

Question 2:

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

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

The midpoint is:

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)

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

correct answer:

b.(19,-2)