Geometry and Trigonometry - Enlargement around a given point-extended

Triangle $ABC$ has vertices $A(2, 2)$, $B(4, 2)$, and $C(2, 5)$. Enlarge the triangle by a scale factor of $k = 2$ using the center of enlargement $(1, 1)$. Find the coordinates of the image $A'B'C'$.

Using the formula $(x', y') = (a + k(x - a), b + k(y - b))$ where $(a, b) = (1, 1)$ and $k = 2$: For $A(2, 2)$: $x' = 1 + 2(2 - 1) = 3$, $y' = 1 + 2(2 - 1) = 3 \Rightarrow A'(3, 3)$ For $B(4, 2)$: $x' = 1 + 2(4 - 1) = 7$, $y' = 1 + 2(2 - 1) = 3 \Rightarrow B'(7, 3)$ For $C(2, 5)$: $x' = 1 + 2(2 - 1) = 3$, $y' = 1 + 2(5 - 1) = 9 \Rightarrow C'(3, 9)$

We calculate the horizontal and vertical distances from the center to each vertex, multiply those distances by the scale factor, and add them back to the center's coordinates.

A square has an area of $25 \text{ cm}^2$. It is enlarged with a scale factor of $k = -3$. Calculate the area of the enlarged image.

$\text{Area of Image} = k^2 \times \text{Area of Object}$ $\text{Area} = (-3)^2 \times 25$ $\text{Area} = 9 \times 25 = 225 \text{ cm}^2$

Even though the scale factor is negative (indicating the image is inverted), the area scale factor is always $k^2$, which is positive.

A point $P(6, 8)$ is mapped to $P'(4, 5)$ by an enlargement with center $(2, 2)$. Determine the scale factor $k$.

Distance from center to $P$ in $x$-direction: $6 - 2 = 4$ Distance from center to $P'$ in $x$-direction: $4 - 2 = 2$ $k = \frac{\text{Image distance}}{\text{Object distance}} = \frac{2}{4} = 0.5$

The scale factor can be found by comparing the displacement of the image from the center to the displacement of the original point from the center.

Using the formula $(x', y') = (a + k(x - a), b + k(y - b))$ with $(a, b) = (1, 2)$ and $k = -0.5$: For $L(3, 4)$: $x' = 1 + (-0.5)(3 - 1) = 1 - 1 = 0$ $y' = 2 + (-0.5)(4 - 2) = 2 - 1 = 1$ $L'(0, 1)$

For $M(5, 4)$: $x' = 1 + (-0.5)(5 - 1) = 1 - 2 = -1$ $y' = 2 + (-0.5)(4 - 2) = 2 - 1 = 1$ $M'(-1, 1)$

For $N(3, 6)$: $x' = 1 + (-0.5)(3 - 1) = 1 - 1 = 0$ $y' = 2 + (-0.5)(6 - 2) = 2 - 2 = 0$ $N'(0, 0)$

Since the scale factor is negative, the image appears on the opposite side of the center (1, 2) and is inverted. Because $|k| < 1$, the image is smaller than the original triangle.

1. Area of Original: $\text{Width} = 3 - 1 = 2$, $\text{Height} = 2 - 1 = 1$. $\text{Area} = 2 \times 1 = 2 \text{ units}^2$.
2. New coordinates: $(1, 1) \to (3, 3)$ $(3, 1) \to (9, 3)$ $(3, 2) \to (9, 6)$ $(1, 2) \to (3, 6)$
3. Area of Image: $\text{Width} = 9 - 3 = 6$, $\text{Height} = 6 - 3 = 3$. $\text{Area} = 6 \times 3 = 18 \text{ units}^2$.
4. Ratio: $\frac{18}{2} = 9$, which is $k^2$ ($3^2 = 9$).

The area increases by the square of the linear scale factor. Here, $k=3$, so the area increases by a factor of 9.