Question

Triangle XYZ is translated 4 units up and 3 units left to yield ΔXYZ. What is the distance between any two corresponding points on ΔXYZ and ΔXYZ′?

Answer 1

Answer:

5 units

Step-by-step explanation:

A point 4 up and 3 over from its original location has moved a distance that can be found using the Pythagorean theorem.

... d² = 4² +3² = 16 +9 = 25

... d = √25 = 5 . . . . units

Answer 2

Answer:

5 units

Step-by-step explanation:

Assuming that any point on the initial triangle XYZ was the origin (0,0).

The distance of between the two points can be calculated by using the Pythagoras Theorem to solve the problem:

$c^2=a^2+b^2$

where $c$ is the hypotenuse and $a$ and $b$ are the legs of the triangle.

$c^2=3^2+4^2\\\\c^2=9+16\\\\c^2=25\\\\c=5$

Therefore, we can conclude that the the distance between any two corresponding points on ΔXYZ and ΔX′Y′Z′  is 5 units.