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Using the transformation T : (x, y) → (x + 2, y + 1), find the distance named. Round the distance to the nearest hundredth of a unit. Complete your work in the space provided or upload a file that can display math symbols if your work requires it.

Find the distance C'A'.

Using the transformation T x y x 2 y 1 find the distance named Round the distance to the nearest hundredth of a unit Complete your work in the space provided or class=

Respuesta :

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C' is located at (0, 3) and A' is located at (2, 1).

d=√(x₂-x₁)²+(y₂-y₁)²
d=√(2-0)²+(1-3)²
d=√(2)²+(-2)²
d=√4+4
d=√8
d=2.83

The distance between points C' and A' is 2.83 units.

Answer:

The distance C'A' is [tex]2\sqrt{2}[/tex] units.

Step-by-step explanation:

From the given graph it is clear that the coordinates of C' and A' are C'(0,3) and A'(2,1).

Distance formula:

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

Using distance formula, the distance between C' and A' is

[tex]C'A'=\sqrt{(2-0)^2+(1-3)^2}[/tex]

On simplification we get

[tex]C'A'=\sqrt{(2)^2+(-2)^2}[/tex]

[tex]C'A'=\sqrt{4+4}[/tex]

[tex]C'A'=\sqrt{8}[/tex]

[tex]C'A'=2\sqrt{2}[/tex]

Therefore the distance C'A' is [tex]2\sqrt{2}[/tex] units.

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