Question

Using the transformation T: (x, y) (x + 2, y + 1), find the distance named. Find the distance A'B'.
(I need help knowing how to work it not just the answer)

Answer 1

we know that

the rule of the translation is

$(x,y)------> (x+2,y+1)$

that means

the translation is $2$ units to the right  and $1$ unit up

Let

$A(0,0)\\B(1,3)$

Step 1

Find the coordinates of A'

$A(0,0)------> A'(0+2,0+1)$

$A(0,0)------> A'(2,1)$

Find the coordinates of B'

$B(1,3)------> B'(1+2,3+1)$

$B(1,3)------> B'(3,4)$

Step 2

Find the distance A'B'

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$A'(2,1)\\B'(3,4)$

substitute in the formula

$d=\sqrt{(4-1)^{2}+(3-2)^{2}}$

$dA'B'=\sqrt{10}\ units$

therefore

the answer is

the distance A'B' is $\sqrt{10}\ units$

Answer 2

The distance A'B' as shown in the diagram is 3.16 units.

Transformation

Transformation is the movement of a point from its initial location to a new location. Types of transformation are rotation, reflection, translation and dilation.

If a point A(x, y) is translated a units right and b unit up, the new point is at A'(x + b, y + b).

The location of the point is at:

• A'(2, 1) and B'(3, 4)

Hence:

• $A'B'=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2} =\sqrt{(4-1)^2+(3-2)^2}=3.16\ units$

The distance A'B' as shown in the diagram is 3.16 units.

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