Exercise 21 Page 202 - 1. Section 4.1 - Core Connections Integrated I, 2013

a Let's start by plotting the given triangle on graph paper.

Next, let's use the function

(x, y) \to (- 1 x, - 1 y)

to find the coordinates of the transformation we want to make.

Point A B C (x, y) (2, 2) (- 2, 2) (8, - 2) (- 1 x, - 1 y) (- 2, - 2) (2, - 2) (- 8, 2)

Now we can graph the transformed triangle.

Notice that multiplying the $x -$ and $y -$ coordinates by $- 1$ corresponds to a rotation about the origin by $18 0^{\circ} .$ A rotation is a rigid motion, which means the shape and size of the preimage has been preserved.

b Let's use the new function to find the coordinates of the points.

Point A B C (x, y) (2, 2) (- 2, 2) (8, - 2) (- 2 x, - 2 y) (- 4, - 4) (4, - 4) (- 16, 4)

Now we can graph the transformed triangle.

Again, we have multiplied the $x -$ and $y -$ coordinates by a negative number, which rotates the figure about the origin by $18 0^{\circ} .$ However, this time the number is not $- 1$ but $- 2,$ which dilates the triangle by a factor of $2 .$ A dilation preserves shape but not size. Therefore, $△ A^{''} B^{''} C^{''}$ has the same shape as $△ A B C,$ but not the same size.