Geometry and Trigonometry - Enlargement by a rational factor-extended

A triangle $ABC$ has coordinates $A(2, 2)$, $B(4, 2)$, and $C(2, 6)$. Find the coordinates of the image $A'B'C'$ after an enlargement with center $(0, 0)$ and a rational scale factor of $k = \frac{3}{2}$.

Multiply each coordinate by $k = 1.5$: $A' = (2 \times 1.5, 2 \times 1.5) = (3, 3)$ $B' = (4 \times 1.5, 2 \times 1.5) = (6, 3)$ $C' = (2 \times 1.5, 6 \times 1.5) = (3, 9)$

Since the center of enlargement is the origin, we apply the mapping $(x, y) \rightarrow (kx, ky)$ directly to each vertex.

A rectangle has an area of $20 \text{ cm}^2$. It undergoes an enlargement with a scale factor of $k = -\frac{1}{2}$. Calculate the area of the resulting image.

$\text{Area of Image} = k^2 \times \text{Area of Object}$ $\text{Area of Image} = (-\frac{1}{2})^2 \times 20$ $\text{Area of Image} = \frac{1}{4} \times 20 = 5 \text{ cm}^2$

Even though the scale factor is negative, the area scale factor is always positive because $k$ is squared ($k^2$). The negative sign indicates the image is inverted, but the size reduction depends only on the magnitude $|\frac{1}{2}|$.

A point $P(5, 8)$ is enlarged to $P'(7, 13)$ from a center of enlargement $C(3, 3)$. Determine the rational scale factor $k$.

Use the vector components from the center: Horizontal distance from $C$ to $P$: $5 - 3 = 2$ Horizontal distance from $C$ to $P'$: $7 - 3 = 4$ $k = \frac{\text{Image Distance}}{\text{Object Distance}} = \frac{4}{2} = 2$ Check with vertical distances: Vertical distance from $C$ to $P$: $8 - 3 = 5$ Vertical distance from $C$ to $P'$: $13 - 3 = 10$ $k = \frac{10}{5} = 2$

The scale factor $k$ is the ratio of the distances from the center to the image and the center to the object. Both the $x$ and $y$ displacements must yield the same $k$.

1. Multiply each coordinate by $k = \frac{3}{4}$: $A' = (4 \times \frac{3}{4}, 4 \times \frac{3}{4}) = (3, 3)$ $B' = (8 \times \frac{3}{4}, 4 \times \frac{3}{4}) = (6, 3)$ $C' = (8 \times \frac{3}{4}, 8 \times \frac{3}{4}) = (6, 6)$ $D' = (4 \times \frac{3}{4}, 8 \times \frac{3}{4}) = (3, 6)$
2. The ratio of the areas is $k^2 = (\frac{3}{4})^2 = \frac{9}{16}$.

Since the center is the origin, we apply the scalar $k$ directly to the coordinates. The resulting square is smaller because $|k| < 1$.

1. The new dimensions are found by taking the absolute value of the scale factor: $\text{Base}' = |k| \times 10 = \frac{2}{5} \times 10 = 4 \text{ units}$ $\text{Height}' = |k| \times 6 = \frac{2}{5} \times 6 = 2.4 \text{ units}$
2. Calculate Area of $T$: $\frac{1}{2} \times 10 \times 6 = 30 \text{ sq units}$
3. Calculate Area of $T'$: $k^2 \times \text{Area of } T = (-\frac{2}{5})^2 \times 30 = \frac{4}{25} \times 30 = 4.8 \text{ sq units}$.

A negative scale factor indicates the image is inverted. Lengths are always positive, so we use $|k|$ for dimensions, but $k^2$ for area transformations.