The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2)

What are the coordinates of Point C'? C'(___,____)

What are the coordinates of Point A? A(____,_____)

Answers:

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Explanation:

I'm assuming you mean to say "scale factor of s = 1/3" since the green triangle is smaller than the red triangle.

Let's start at the point (4,2). This is the center of dilation. This fixed point does not move or change as you apply any dilation (as long as you keep this as the center of course). Every other point will move. For instance, the point (10,14) moves to (6,6) after applying the dilation.

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The x coordinate of the center point (4,2) is x = 4.

The x coordinate of the point (10,14) is x = 10.

Going from (4,2) to (10,14) will have us move a horizontal distance of 10-4 = 6 units.

Take 1/3 of this horizontal distance and we get (1/3)*6 = 2. This means that if we started at the center point (4,2) and move 2 units to the right, we will have the x coordinate to be x = 4+2 = 6, which matches up with the x coordinate of the point (6,6)

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We'll do the same sort of thing with the y coordinates

The point (4,2) has the y coordinate y = 2

The point (10,14) has the y coordinate y = 14

The vertical distance between those points is 14-2 = 12 units

One third of this is (1/3)*12 = 4

If we started at (4,2) and moved up 4 units, then the y coordinate goes from y = 2 to y = 2+4 = 6. This also matches (6,6)

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Hopefully by now you have a sense of how this all works. Let's use these tricks to find the location of point A.

The horizontal distance from (4,2) to (8,4) is 8-4 = 4 units. This triples to 4*3 = 12 units. We triple here because we're going in reverse of the process described in the previous two sections. If (8,4) and A swapped places, then we would take one third instead.

We'll use that 12 to count out 12 horizontal spaces from (4,2) to get to (16,2). I added 12 to the x coordinate of (4,2).

Repeat the process for the y values. The vertical distance from (4,2) to (8,4) is 2 units, which triples to 6. Add 6 onto the y coordinate of (16,2) and we arrive at (16, 8) which is the location of point A.

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Now let's find the location of point C'

The horizontal distance from (-2,11) to (4,2) is 6 units because |-2-4| = 6. Multiply this by 1/3 and we get (1/3)*6 = 2

To get to the location point C', we need to start at (4,2) and move to the left 2 units. That means (4,2) will move to (2,2) as a temporary location.

We apply the same idea for the y coordinates. The vertical distance from (4,2) to (-2,11) is 9 units. This leads to (1/3)*9 = 3 units being the vertical distance from (4,2) to the location of C'. We'll add 3 to the y coordinate of (2,2) to arrive at (2,5) which is the location of C'.

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