Segment AB is dilated from the origin to create segment A'B' at A' (0, 6) and B' (6, 9). What scale

10 answers
Question:

Segment AB is dilated from the origin to create segment A'B' at A' (0, 6) and B' (6, 9). What scale factor was segment AB dilated by? A) 1/2. B) 2. C) 3. D) 4.​ {Please look at the comments for the answer.}

Answers

2 is the answer of the question

3.

Step-by-step explanation:

When a figure with vertices (x,y) is dilated from the origin by a scale factor 'k' .

Then , the coordinates of the image is given by :-

Given : Segment AB is dilated from the origin to create segment A prime B prime at A' (0, 6) and B' (6, 9).

From the graph , coordinates of AB are A(0,2) and B(2,3).

Let k be the scale factor then ,

Since A' is the image of point A.

Then,

Hence, the scale factor was 3.

3

Step-by-step explanation:

From (0,2) to (0,6) is a factor of 3

From (2,3) to (6,9) is a factor of 3

4

Step-by-step explanation:

I took the test and got it right.

answer:

step-by-step explanation:

ab & op

look at photo

[tex]Segment ab is dilated from the origin to create segment a prime b prime at a' (0, 6) and b' (6, 9).[/tex]

Scale factor is 3.

Step-by-step explanation:

Given: The Line Segment AB is dilated and a line segment A'B' is created.

A' (0, 6) and B' (6, 9).

As per the given graph , we can easily see that the coordinates of A and B are

A(0, 2) and B(2, 3)

Please refer to attached graph for the coordinates of A', B' and A, B.

The distance formula is given as:

[tex]D =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

D is the distance between two coordinates [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex].

So, the distance AB = [tex]\sqrt{(2-0)^2+(3-2)^2} = \sqrt{5}[/tex]

Similary, distance A'B' = [tex]\sqrt{(6-0)^2+(9-6)^2} = \sqrt{36 + 9}=\sqrt{45} = 3\sqrt{5}[/tex]

We can clearly see that A'B' is three times the distance AB.

So, The scale factor 3.


[tex]segment AB is dilated from the origin to create segment A prime B prime at A' (0, 6) and B' (6, 9).[/tex]

The answer is "Option C"

Step-by-step explanation:

Given:

point A' and B'  are:  (0, 6)  (6, 9)

To above points we calculate point AB that is (0,2) and (2,3)

Distance formula:

[tex]\bold{D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}\\[/tex]

calcuate AB point distance:

[tex]x_1=0\\y_1=2\\x_2=2\\y_2=3[/tex]

[tex]D=\sqrt{(2-0)^2+(3-2)^2}[/tex]

    [tex]=\sqrt{2^2+1^2}\\\\=\sqrt{4+1}\\\\=\sqrt{5}\\[/tex]

calculating A'B' point distance:

[tex]x_1=0\\y_1=6\\x_2=6\\y_2=9\\[/tex]

[tex]D=\sqrt{(6-0)^2+(9-6)^2}[/tex]

   [tex]= \sqrt{(6)^2+(3)^2}\\\\= \sqrt{36+9}\\\\= \sqrt{45}\\\\= \sqrt{3\times 3\times 5}\\\\= 3\sqrt{5}\\\\[/tex]

If we divide:

[tex]=\frac{3\sqrt{5}}{\sqrt{5}}\\\\=3[/tex]

The final answer is "3".

the scale factor is definently 3

Step-by-step explanation:

C) 3

Work Shown:

Point A is located at (0,2). It moves to A ' (0,6). Focus on the y coordinates. We go from y = 2 to y = 6. This is a jump of multiplying by 3. We have multiplied every coordinate of A by 3 to get corresponding coordinates for A'.

Point B will be treated the same way. It starts off at B = (2,3) and moves to B ' = (6,9) after multiplying both coordinates by the same scale factor 3.

The length of AB will also be tripled

length of A'B' = 3*(length of AB)

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