Geometry and Trigonometry - Geometric transformations: translation, reflection, rotation, and enlargement

Triangle $ABC$ has vertices $A(1, 2)$, $B(4, 2)$, and $C(1, 6)$. Apply a translation using the vector $\begin{pmatrix} -3 \\ 4 \end{pmatrix}$ and find the new coordinates of the vertices.

1. Identify the $x$-shift and $y$-shift from the vector: $x = -3$, $y = 4$.
2. Add the shifts to each vertex: $A'(1 + (-3), 2 + 4) = (-2, 6)$ $B'(4 + (-3), 2 + 4) = (1, 6)$ $C'(1 + (-3), 6 + 4) = (-2, 10)$
3. The translated vertices are $A'(-2, 6)$, $B'(1, 6)$, and $C'(-2, 10)$.

To translate a point, we add the top value of the vector to the $x$-coordinate and the bottom value to the $y$-coordinate.

A square has a vertex at $P(2, 3)$. It undergoes an enlargement with a scale factor $k = 3$ centered at the origin $(0, 0)$. Determine the coordinates of the image vertex $P'$. If the original square had an area of $5 \text{ cm}^2$, what is the area of the enlarged square?

1. For the coordinates: Use the rule $(x, y) \rightarrow (kx, ky)$. $P'(3 \times 2, 3 \times 3) = (6, 9)$.
2. For the area: The area of an enlarged shape is the original area multiplied by the scale factor squared ($k^2$). $\text{New Area} = 5 \times 3^2 = 5 \times 9 = 45 \text{ cm}^2$.

When the center of enlargement is the origin, we simply multiply coordinates by $k$. Note that while side lengths increase by $k$, area increases by $k^2$.

The rule for reflection in the line $y = -x$ is $(x, y) \rightarrow (-y, -x)$. Applying this to the vertices: $L(1, 1) \rightarrow L'(-1, -1)$ $M(3, 1) \rightarrow M'(-1, -3)$ $N(1, 4) \rightarrow N'(-4, -1)$ Therefore, the new vertices are $L'(-1, -1)$, $M'(-1, -3)$, and $N'(-4, -1)$.

To reflect a point across the line $y = -x$, we swap the $x$ and $y$ coordinates and then multiply both by $-1$. This results in a mirror image located in the third quadrant if the original was in the first.

1. Find the vector from the center of rotation $(1, 1)$ to point $A(2, 2)$: $\vec{CA} = \begin{pmatrix} 2 - 1 \\ 2 - 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
2. A $180^{\circ}$ rotation reverses the vector: $\vec{CA'} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}$
3. Add this vector to the center of rotation $(1, 1)$: $A' = (1 + (-1), 1 + (-1)) = (0, 0)$ The coordinates of $A'$ are $(0, 0)$.

A rotation of $180^{\circ}$ about any point $(h, k)$ is equivalent to a point reflection. The formula for the image is $P'(2h - x, 2k - y)$. For point $A(2, 2)$ and center $(1, 1)$, we get $(2(1)-2, 2(1)-2) = (0, 0)$.