Exercise 36 Page 213

We want to think about how to dilate a triangle and find its center of dilation. In general, the process takes the following steps. In order to think about this process, we will consider two example scenarios with arbitrarily chosen vertices and scale factors.

First Example

The center of dilation is the origin.

Finding the New Triangle's Coordinates

We will choose the vertices of $△ A B C$ to be $A (1, 1),$ $B (1, 3)$ and $C (3, 1) .$

Now we can draw an image of the triangle that is a dilation of

△ A B C .

For our example, we will use a scale factor of

k = 2 .

When the center of dilation is the origin, we can use the Coordinate Rule to find the image's coordinates. Let's multiply each of the coordinates by

k .

(x, y) A (1, 1) B (1, 3) C (3, 1) (2 x, 2 y) (2 (1), 2 (1)) (2 (1), 2 (3)) (2 (3), 2 (1)) (x^{'}, y^{'}) A^{'} (2, 2) B^{'} (2, 6) C^{'} (6, 2)

Second Example

The center of dilation is unknown and it is not the origin.

Finding the New Triangle's Coordinates

For this example, let's start with the same coordinates as in the first example.

A (1, 1), B (1, 3), and C (3, 1)

This time, we will not begin with a scale factor. Instead we will choose the points based on a new location on the coordinate plane and keeping the existing proportions of the side lengths. The length of the base and the height of

△ A B C

are both

2 .

We can choose

3

to be the new lengths of the base and the height for

△ A^{'} B^{'} C .

△ A B C A B = 2 A C = 2 △ A^{'} B^{'} C A^{'} B^{'} = 3 A^{'} C^{'} = 3

In this case, the image will be bigger than the preimage so it is considered an enlargement — if the dilation make the figure smaller, it would be a reduction. Now we can place the image anywhere we want in the coordinate system. We will start with the vertex

A^{'}

and place it at

(3, - 1) .

Next, the vertex $B^{'}$ should be placed $3$ units above $A^{'}$ and $C^{'}$ should be placed $3$ units to the right.

Finding the Center of Dilation

The last step is to find the center of dilation. It can be found by drawing lines from the vertices of $△ A^{'} B^{'} C^{'}$ through the corresponding vertices of $△ A B C .$ The point where all three lines cross each other will be the center $P .$

The center of the dilation is therefore $P (- 3, 5) .$