Question

for exercises 3 and 4, find the coordinates of the vertices of each image.

1. ( d_{0,75}(\triangle abc) ), given ( a(4, -3) ), ( b(6, 1) ), ( c(10, -1) )
2. ( d_{1.5}(\triangle xyz) ), given ( x(3, 0) ), ( y(4, 2) ), ( z(6, -2) )
3. ( d_k(\triangle abc) ) has a perimeter of 100 units and an area of 625 units(^2).

a. what is the perimeter of ( \triangle abc )?
b. what is the area of ( \triangle abc )?

Explanation:

Step1: Dilate point A by 0.75

Multiply each coordinate of $A(4, -3)$ by 0.75:
$A'(4\times0.75, -3\times0.75)=(3, -2.25)$

Step2: Dilate point B by 0.75

Multiply each coordinate of $B(6, 1)$ by 0.75:
$B'(6\times0.75, 1\times0.75)=(4.5, 0.75)$

Step3: Dilate point C by 0.75

Multiply each coordinate of $C(10, -1)$ by 0.75:
$C'(10\times0.75, -1\times0.75)=(7.5, -0.75)$

Step4: Dilate point X by 1.5

Multiply each coordinate of $X(3, 0)$ by 1.5:
$X'(3\times1.5, 0\times1.5)=(4.5, 0)$

Step5: Dilate point Y by 1.5

Multiply each coordinate of $Y(4, 2)$ by 1.5:
$Y'(4\times1.5, 2\times1.5)=(6, 3)$

Step6: Dilate point Z by 1.5

Multiply each coordinate of $Z(6, -2)$ by 1.5:
$Z'(6\times1.5, -2\times1.5)=(9, -3)$

Step7: Find scale factor k (perimeter)

Perimeter scales by k: $k=\frac{\text{Image Perimeter}}{\text{Pre-image Perimeter}}=\frac{625}{100}=6.25$

Step8: Find area of image $\triangle ABC$

Area scales by $k^2$: $\text{Area}=625\times(6.25)^2=625\times39.0625=24414.0625$

Answer:

1. $A'(3, -2.25)$, $B'(4.5, 0.75)$, $C'(7.5, -0.75)$
2. $X'(4.5, 0)$, $Y'(6, 3)$, $Z'(9, -3)$

5a. 625 units
5b. 24414.0625 units²