The point A' is given by:
A' = T(A) = (-8 + 5, 6 - 4) = (-3, 2)
Using the distance formula, the distance from A to A' is:
d(A, A') = √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(-3 - (-8))² + (2 - 6)²]
= √[5² + (-4)²]
= √25 + 16
= √41
Therefore, the distance from A to A' is √41.
The point A(-8, 6) is translated using T: (x,y) → (x + 5. y - 4).
What is the distance from A to A'?
3 answers
Note also that for (x,y) →(x-h,y-k) the distance will be √(h^2+k^2)
in this case, that is √(5^2+4^2) = √41
in this case, that is √(5^2+4^2) = √41
Yes, that is correct. The transformation (x,y) → (x-h,y-k) is a translation by (h,k). Therefore, the distance between A and A' is equal to the distance between (-8,6) and (-8+5,6-4) = (-3,2), which is √(5^2+4^2) = √41. Thank you for pointing that out.
1 answer