Using the transformation T: (x, y) (x + 2, y + 1), find the distance named. Find the distance A'B'.

Solution:

Given, T: (x, y) (x + 2, y + 1) --- (1)

We have to find the distance using translation.

We know that the rule of the translation is

(x, y) → (x + a, y + b) --- (2)

Comparing (1) and (2)

The translation is

a = 2 ( 2 units right)

b = 1 (1 unit up)

From the figure, the coordinates of

A = (0, 0), B = (1, 3)

Using translation, the coordinates of

A’ = (0 + 2, 0 + 1) = (2, 1)

B’ = (1 + 2, 3 + 1) = (3, 4)

Now, finding the distance A’B’ using the distance formula

Distance = \(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

A’B’ = \(\sqrt{(3-2)^{2}+(4-1)^{2}}\)

= \(\sqrt{(1)^{2}+(3)^{2}}\)

= \(\sqrt{1+9}\)

= \(\sqrt{10}\)

Therefore, the distance A’B’ is \(\sqrt{10}\)units.

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Using the transformation T: (x, y) (x + 2, y + 1), find the distance named. Find the distance A'B'.

Summary:

Using the transformation T: (x, y) (x + 2, y + 1), the distance A’B’ is \(\sqrt{10}\)units.