M - Translations and Area Lesson

Translations and Area

Matrices have geometrical applications too!

Let's say we have a parallelogram with the following vertices: A(-1, -2), B (4, -2), C (6, 1), D(1, 1)

We could put those vertices in a matrix: equation image indicator $LaTeX: P=\begin{bmatrix} -1 & 4 & 6 & 1\\ -2 & -2 & 1 & 1 \end{bmatrix}$ $P=\begin{bmatrix} -1 & 4 & 6 & 1\\ -2 & -2 & 1 & 1 \end{bmatrix}$

Then, what do you think would happen to the parallelogram if we did this: equation image indicator $LaTeX: P=\begin{bmatrix} -1 & 4 & 6 & 1\\ -2 & -2 & 1 & 1 \end{bmatrix} + P=\begin{bmatrix} -1 & -1 & -1 & -1\\ 2 & 2 & 2 & 2 \end{bmatrix}$ $P=\begin{bmatrix} -1 & 4 & 6 & 1\\ -2 & -2 & 1 & 1 \end{bmatrix} + P=\begin{bmatrix} -1 & -1 & -1 & -1\\ 2 & 2 & 2 & 2 \end{bmatrix}$

The resulting matrix would be: equation image indicator $LaTeX: P=\begin{bmatrix} -2 & 3 & 5 & 0\\ 0 & 0 & 3 & 3 \end{bmatrix}$ $P=\begin{bmatrix} -2 & 3 & 5 & 0\\ 0 & 0 & 3 & 3 \end{bmatrix}$

image of parallelogram ABCD plotted twice on graph

Let's say I have a set of coordinates like the parallelogram: A(-1, -2), B (4, -2), C (6, 1), D(1, 1)

Since I know that when you multiply matrix the "number of columns in the first matrix has to match with the number of rows of the second matrix". So the transformation matrix will be in the front. Because the transformation matrix is a 2x2 matrix. When we put the coordinates of a polygon as a matrix, the number of rows will always be 2. Therefore, when multiplying matrix for transformation. The transformation matrix must come first.

equation image indicator $LaTeX: \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} -1 & 4 & 6 & 1 \\ -2 & -2 & 1 & 1 \end{bmatrix}$ $\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} -1 & 4 & 6 & 1 \\ -2 & -2 & 1 & 1 \end{bmatrix}$

equation image indicator $LaTeX: \begin{bmatrix} (-1)(-1)+0(-2) & (-1)(4)+0(-2) & (-1)(6)+(0)(1) & (-1)(1)+(0)(1)) \\ (0)(-1)+(1)(2) & (0)(4)+(1)(-2) & (0)(6)+(1)(1) & (0)(1)+(1)(1) \end{bmatrix}$ $\begin{bmatrix} (-1)(-1)+0(-2) & (-1)(4)+0(-2) & (-1)(6)+(0)(1) & (-1)(1)+(0)(1)) \\ (0)(-1)+(1)(2) & (0)(4)+(1)(-2) & (0)(6)+(1)(1) & (0)(1)+(1)(1) \end{bmatrix}$

equation image indicator $LaTeX: = \begin{bmatrix} 1 & -4 & -6 & -1 \\ -2 & -2 & 1 & 1 \end{bmatrix}$ $= \begin{bmatrix} 1 & -4 & -6 & -1 \\ -2 & -2 & 1 & 1 \end{bmatrix}$

This means this is a transformation on a reflection over the y-axis. Notice the values of the coordinates changed sign.

Watch the video to see how to use the transformation matrix to translate a triangle in the coordinate plane.

1. What transformation matrix represents the translation from equation image indicator $LaTeX: \bigtriangleup ABC\longrightarrow\bigtriangleup A'B'C'$ $\bigtriangleup ABC\longrightarrow\bigtriangleup A'B'C'$ ?

Solution: $LaTeX: \begin{bmatrix} -3 & -3 & -3\\ 1 & 1 & 1 \end{bmatrix}$ $\begin{bmatrix} -3 & -3 & -3\\ 1 & 1 & 1 \end{bmatrix}$

image of triangle ABC plotted on graph

2. If T is a vertices matrix, what does equation image indicator $LaTeX: T+ \begin{bmatrix} 5 & 5 & 5 & 5 \\ -4 & -4 & -4 & -4 \end{bmatrix}$ $T+ \begin{bmatrix} 5 & 5 & 5 & 5 \\ -4 & -4 & -4 & -4 \end{bmatrix}$ represent?

Solution: Translated right 5 and down 4

3. If T is a vertices matrix, what does equation image indicator $LaTeX: T+ \begin{bmatrix} -3 & -3 & -3 & -3 \\ 7 & 7 & 7 & 7 \end{bmatrix}$ $T+ \begin{bmatrix} -3 & -3 & -3 & -3 \\ 7 & 7 & 7 & 7 \end{bmatrix}$ represent?

Solution: Translated left 3 and up 7

Area of a Parallelogram

We can also use the determinant of a matrix to find the area of a parallelogram if it is in a specific form.

Let's say we have a parallelogram with vertices (0, 0) (a, b) (c, d) and (a + c, b + d). This looks complicated but really it's just saying that one of the points is the sum of the other non-zero x's and y's.

We can write a matrix with a, b, c, and d: equation image indicator $LaTeX: \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , then the area of the parallelogram is the absolute value of the determinant: A = |ad - bc|

Try these to see if you've got it:

If you have a parallelogram with vertices A(0, 0) B(2, 4) C(3, 9) D(5, 13), find the area.
- Solution: 6
If you have a parallelogram with vertices A(0, 0) B(-3, 4) C(-1, 3) D(-4, 7), find the area.
- Solution: 5
If you have a parallelogram with vertices A(0, 0) B(-1, -4) C(4, 2) D(3, -2), find the area.
- Solution: 14

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