Question

let (a : (0,0)), (b : (1,0)), and (c : (0,2)) be the vertices of triangle (delta abc) in (mathbb{r}^2). find the coordinates of points (a), (b), and (c) in (mathbb{r}^2) such that (delta abc) is the image of (delta abc) under the composition of dilations (h_{q,3} circ h_{p,2}), where (p : (-1,1)) and (q : (3,2)).

Explanation:

⚡ Using what you learned: Compositions of Transformations · 🆕 New: Dilations from Non-Origin Centers
Scaling points relative to a custom center

Step 1: Understand the Dilation Formula

A dilation \( H_{C, k} \) centered at a point \( C(x_c, y_c) \) with a scale factor \( k \) maps any point \( X(x, y) \) to a new point \( X'(x', y') \) using the vector relation:

$$ \vec{CX'} = k \cdot \vec{CX} $$

In coordinate form, this is:

$$ x' = x_c + k(x - x_c) $$

$$ y' = y_c + k(y - y_c) $$

We need to apply the composition \( H_{Q,3} \circ H_{P,2} \) to the vertices \( A(0,0) \), \( B(1,0) \), and \( C(0,2) \). This means we first apply \( H_{P,2} \) with center \( P(-1, 1) \) and scale factor \( 2 \), and then apply \( H_{Q,3} \) with center \( Q(3, 2) \) and scale factor \( 3 \) to the resulting points.

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Step 2: Apply the First Dilation \( H_{P,2} \)

The center is \( P(-1, 1) \) and the scale factor is \( k_1 = 2 \). The formula for the intermediate points \( (x', y') \) is:

$$ x' = -1 + 2(x + 1) = 2x + 1 $$

$$ y' = 1 + 2(y - 1) = 2y - 1 $$

Let's calculate the intermediate coordinates \( A' \), \( B' \), and \( C' \):

• For \( A(0,0) \):

$$ x'_A = 2(0) + 1 = 1 $$

$$ y'_A = 2(0) - 1 = -1 $$

$$ A' = (1, -1) $$

• For \( B(1,0) \):

$$ x'_B = 2(1) + 1 = 3 $$

$$ y'_B = 2(0) - 1 = -1 $$

$$ B' = (3, -1) $$

• For \( C(0,2) \):

$$ x'_C = 2(0) + 1 = 1 $$

$$ y'_C = 2(2) - 1 = 3 $$

$$ C' = (1, 3) $$

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Step 3: Apply the Second Dilation \( H_{Q,3} \)

The center is \( Q(3, 2) \) and the scale factor is \( k_2 = 3 \). The formula to find the final coordinates \( (x'', y'') \) from the intermediate coordinates \( (x', y') \) is:

$$ x'' = 3 + 3(x' - 3) = 3x' - 6 $$

$$ y'' = 2 + 3(y' - 2) = 3y' - 4 $$

Let's calculate the final coordinates \( A'' \), \( B'' \), and \( C'' \):

• For \( A'(1, -1) \):

$$ x''_A = 3(1) - 6 = -3 $$

$$ y''_A = 3(-1) - 4 = -7 $$

$$ A'' = (-3, -7) $$

• For \( B'(3, -1) \):

$$ x''_B = 3(3) - 6 = 3 $$

$$ y''_B = 3(-1) - 4 = -7 $$

$$ B'' = (3, -7) $$

• For \( C'(1, 3) \):

$$ x''_C = 3(1) - 6 = -3 $$

$$ y''_C = 3(3) - 4 = 5 $$

$$ C'' = (-3, 5) $$

Answer:

The coordinates of the image points are:

$$ A'' = (-3, -7) $$

$$ B'' = (3, -7) $$

$$ C'' = (-3, 5) $$