Question

A line segment has endpoints at `(2 ,7 )` and `(5 ,4 )`. The line segment is dilated by a factor of `3` around `(4 ,3 )`. What are the new endpoints and length of the line segment?

Answer 1

Please read the explanation.

Explanation:

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Given:

Scale Factor for dilation is `3`.

Useful observations involving Dilation:

Isometry refers to a linear transformation which preserves the length.

Dilation is NOT an isometry. It creates similar figures only.

Here `bar (AB)` is the pre-image and after dilation, `bar (A'B')` is called the image.

The absolute value of the scale factor (k),

with the constraint `0 < k < 1,`

reduces the line segment `bar (AB)`,

enlarges if otherwise.

Each point on the line segment `bar (AB)` will get 3 times as far from the Center of Dilation, `(4,3)` since the scale factor is `3`.

Dilation preserves the angle of measure.

Note that the pre-image and the image are parallel.

Observe that the points (center of dilation `color(red)C` , A and A') collinear.

And, the points (C, B and B') are also collinear.

`bar (AB) |\| bar (A'B')`, since we have congruent corresponding angles.

Also, from `C(4,3)`, move up 4 units on the y-axis, and 2 units left on the x-axis to reach the end-point `A(2,7)`.

Move (4 x 3 = 12 units) up on the y-axis, and (2 x 3 = 6 units) left on the x-axis tor reach the end-point of `A'B'(-2, 15)`.

Similarly,

from `C(4,3)`, move one unit up on the y-axis and one unit right on the x-axis, to reach point `B(5,4)`.

From `C(4,3)`, move (1 x 3 = 3 units) on the y-axis, (1 x 3 = 3 units) to the right on the x-axis, to reach the point `B'(7,6)`.

New end-points: `A'(-2, 15) and B'(7,6)`

Find the length of `bar (A'B')`, using distance formula:

`color(blue)(D = sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`rArr D=sqrt[(7-(-2)^2)+(6-15)^2)`

`rArr D=sqrt(9^2+(-9)^2)`

`rArr D=sqrt(162)`

`rArr D~~ 12.72792`

`bar (A'B')~~"12.73 units"`

Hope this solution is helpful.