The point A(-8, 6) is translated using T: (x,y) → (x + 5. y - 4).
What is the distance from A to A'?
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.
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
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.
To find the distance between point A and its image A' after translation, we can use the distance formula. The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Here, (x1, y1) represents the coordinates of point A (-8, 6), and (x2, y2) represents the coordinates of point A' after translation.
According to the translation rule T: (x, y) → (x + 5, y - 4), the new coordinates of A' would be obtained by adding 5 to the x-coordinate and subtracting 4 from the y-coordinate.
So the coordinates of A' can be calculated as follows:
x2 = -8 + 5 = -3
y2 = 6 - 4 = 2
Now we can substitute the values into the distance formula:
d = sqrt((-3 - (-8))^2 + (2 - 6)^2)
= sqrt(5^2 + (-4)^2)
= sqrt(25 + 16)
= sqrt(41)
Therefore, the distance from A to A' is sqrt(41) or approximately 6.40 units.