Question

Triangle xyz is translated 4 units up and 3 units left to yield triangle x'y'z'. What is the distance between any two corresponding points

Answer 1

5 units is the exact answer. I hope it is the exact.

Answer 2

Answer:

The distance between any two corresponding points is 5 unit.

Step-by-step explanation:

Let (x,y) represents the coordinates of the triangle xyz,

Thus, after translation translated 4 units up and 3 units left,

$(x,y)\rightarrow (x+3,y+4)$

That is, if triangle x'y'z' is the resultant figure after translation,

Then (x+3,y+4) represents the coordinates of triangle x'y'z',

By the distance formula,

The distance between any two corresponding points of triangle xyz and x'y'z',

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

$=\sqrt{3^2+4^2}$

$=\sqrt{9+16}$

$=\sqrt{25}$

$=5\text{ unit}$