Question

Please help The dilation DO,2.5 is applied to triangle ABC below. Use the dilation of triangle ABC and its image to prove that the dilation takes a line passing through the center of the dilation to the same line.

Answer 1

If you dilate triangle ABC by 2.5, then the coordinates will change:

from (1, 2) to (2.5, 5)

from (1, -1) to (2.5, -2.5)

from (-2, 2) to (-5, 5)

If you graph the new coordinates, you can see that line AC and line A'C' are on top of each other, which proves that the dilation takes a line passing through the center of the dilation to the same line.

You can also do this without graphing. Just find the equation of each line. For AC, first find the slope by doing:

-1 - 2 = -3 and 1- (-2) = 3. Next do -3 divided by 3 = -1. Then set up the equation:

y - 2 = -1 (x + 2) -->  

y - 2 = -x - 2 -->  

y = -x                                                                

For A'C'

to find the slope:

-2.5 - 5 = -7.5 and 2.5 - (-5) = 7.5. Next do -7.5 divided by 7.5 = -1. Then set up the equation:

y - 5 = -1 (x + 5) -->

y - 5 = -x - 5 -->

y = -x  

You can see that the two equations are the same, so the lines are the same.

Answer 2

A dilation $D_{O,\,k}$ is a dilation with a scale factor, k centered at the origin.

A scale factor, k means that the distance of the image from the the center of dilation is k times the distance of the pre-image from the center of dilation and the size of the image is k times the size of the pre-image.

Given triangle ABC with vertices A(-2, 2), B(1, 2) and C(1, -1), a dilation $D_{O,\,k}$ will result in the image with vertices A'2.5(-2, 2) = A'(-5, 5), B'2.5(1, 2) = B'(2.5, 5) and C'2.5(1, -1) = C'(2.5, -2.5)

Now, consider line AC, the equation of the line is given by

$\frac{y-2}{x-(-2)} = \frac{-1-2}{1-(-2)} \\ \\ \Rightarrow \frac{y-2}{x+2} = \frac{-3}{1+2} = \frac{-3}{3} =-1 \\ \\ \Rightarrow y-2=-(x+2)=-x-2 \\ \\ \Rightarrow y=-x$

Notice that line y = -x is a line passing through the origin which is the center of dilation.

Consider line A'C', the equation of the line is given by

$\frac{y-5}{x-(-5)} = \frac{-2.5-5}{2.5-(-5)} \\ \\ \frac{y-5}{x+5} = \frac{-7.5}{2.5+5} = \frac{-7.5}{7.5} =-1 \\ \\ \Rightarrow y-5=-(x+5)=-x-5 \\ \\ \Rightarrow y=-x$

As an be seen the line AC and the line A'C' is the same line.