1
Esteban drew Triangle JKL on a coordinate plane with J (−3,5), K (−1,−4), and L (2,4). Then he drew J'K'L', the results of the dilation were (x, y) --> (
\frac{2}{3}
3
2
x,
\frac{2}{3}
3
2
y) What are the coordinates of J'?
assuming you meant (x,y)→(2/3 x, 2/3 y) then
(-3,5)→(-2 , 10/3)
If not, then maybe you can type it in a more readable format.
(-2 1/3, 3 1/3)
Joke mode on:
Well, it seems like Esteban's Triangle JKL went through some kind of beauty treatment called "dilation". I guess it wanted to enhance its coordinates. Now it has a new friend called J'! Let's see where it ended up:
Considering the dilation equation (x, y) --> (2/3 * x, 2/3 * y), let's apply it to the coordinates of J:
J' = (2/3 * (-3), 2/3 * 5)
= (-2, 10/3)
So, the coordinates of J' are approximately (-2, 3.33). Guess J really wanted to move closer to that funny-looking clown over there!
To find the coordinates of J' after the dilation, you can multiply the x-coordinate and y-coordinate of J by 2/3.
Let's calculate the coordinates of J'.
x-coordinate of J':
x' = (-3) * (2/3)
x' = (-6/3)
x' = -2
y-coordinate of J':
y' = 5 * (2/3)
y' = 10/3
Therefore, the coordinates of J' are (-2, 10/3).
To find the coordinates of J', we need to apply the given dilation to the coordinates of J.
Given dilation:
(x, y) -> (2/3x, 2/3y)
Coordinates of J:
J (-3, 5)
Let's apply the dilation to the x-coordinate first:
x-coordinate of J' = (2/3) * (-3) = -2
Now, let's apply the dilation to the y-coordinate:
y-coordinate of J' = (2/3) * 5 = 10/3 or 3.33 (rounded to two decimal places)
Therefore, the coordinates of J' are (-2, 3.33).