Triangle DEF has vertices D(-2,1),E(2,4), and F(3,1) and is dilated by a factor pf 3 using the point (1,1) as the point of dilation. The dilated triangle is names D'E'F'. Using the information to determine the location of D' and F'
D = (1,1) + (-3,0)
E = (1,1) + (1,3)
F = (1,1) + (2,0)
that means
D' = (1,1) + 3*(-3,0) = (1,1)+(-9,0) = (-8,1)
E' = (1,1) + 3*(1,3) = (1,1)+(3,9) = (4,10)
F' = (1,1) + 3*(2,0) = (1,1)+(6,0) = (7,1)
To find the coordinates of the dilated vertices D' and F', we can use the dilation formula. Here's how you can do it:
1. Find the distance between the point of dilation (1,1) and the original vertex D(-2,1) using the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance = sqrt((1 - (-2))^2 + (1 - 1)^2)
= sqrt(3^2 + 0^2)
= sqrt(9 + 0)
= sqrt(9)
= 3
2. Multiply the x and y coordinates of D(-2,1) by the dilation factor, which is 3, since we are dilating by a factor of 3:
D' = (3 * -2, 3 * 1)
= (-6, 3)
So, the coordinates of D' are (-6, 3).
3. Similarly, find the distance between the point of dilation (1,1) and the original vertex F(3,1):
Distance = sqrt((3 - 1)^2 + (1 - 1)^2)
= sqrt(2^2 + 0^2)
= sqrt(4 + 0)
= sqrt(4)
= 2
4. Multiply the x and y coordinates of F(3,1) by the dilation factor, which is 3:
F' = (3 * 3, 3 * 1)
= (9, 3)
So, the coordinates of F' are (9, 3).
Therefore, the coordinates of the dilated vertices D' and F' are (-6, 3) and (9, 3), respectively.
To determine the location of D' and F' after dilating the triangle DEF by a factor of 3 using the point (1,1) as the center of dilation, follow these steps:
1. Find the distance between the center of dilation (1,1) and vertex D(-2,1):
- Δx = 1 - (-2) = 3
- Δy = 1 - 1 = 0
2. Multiply the distance Δx and Δy by the dilation factor of 3:
- Δx' = 3 * 3 = 9
- Δy' = 0 * 3 = 0
3. Add the transformed distances to the coordinates of the center of dilation (1,1) to find the new coordinates of D':
- x' = 1 + Δx' = 1 + 9 = 10
- y' = 1 + Δy' = 1 + 0 = 1
Therefore, the new coordinate of D' is D'(10,1).
4. Repeat steps 1-3 for vertex F(3,1):
- Δx = 1 - 3 = -2
- Δy = 1 - 1 = 0
- Δx' = -2 * 3 = -6
- Δy' = 0 * 3 = 0
- x' = 1 + Δx' = 1 - 6 = -5
- y' = 1 + Δy' = 1 + 0 = 1
Therefore, the new coordinate of F' is F'(-5,1).