# 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?

Apr 11, 2018

#### Explanation:

$\text{ }$
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 $\overline{A B}$ is the pre-image and after dilation, $\overline{A ' B '}$ is called the image.

The absolute value of the scale factor (k),

with the constraint $0 < k < 1 ,$

reduces the line segment $\overline{A B}$,

enlarges if otherwise.

Each point on the line segment $\overline{A B}$ will get 3 times as far from the Center of Dilation, $\left(4 , 3\right)$ 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 $\textcolor{red}{C}$ , A and A') collinear.

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

$\overline{A B} | \setminus | \overline{A ' B '}$, since we have congruent corresponding angles.

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

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 $A ' B ' \left(- 2 , 15\right)$.

Similarly,

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

From $C \left(4 , 3\right)$, 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 $B ' \left(7 , 6\right)$.

New end-points: $A ' \left(- 2 , 15\right) \mathmr{and} B ' \left(7 , 6\right)$

Find the length of $\overline{A ' B '}$, using distance formula:

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

$\Rightarrow D = \sqrt{\left(7 - {\left(- 2\right)}^{2}\right) + {\left(6 - 15\right)}^{2}}$

$\Rightarrow D = \sqrt{{9}^{2} + {\left(- 9\right)}^{2}}$

$\Rightarrow D = \sqrt{162}$

$\Rightarrow D \approx 12.72792$

$\overline{A ' B '} \approx \text{12.73 units}$