Question

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

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

You would use the distance formula using the coordinates of A' and B'
A' (2,1)
B' (3,4)
D=square root of [(x-x)^2 + (y-y)^2]
D=sq rt [(3-2)^2+(4-1)^2]
D=sq rt (1^2 + 3^2)
D=sq rt (1+9)
D=square root of 10