Question

Triangle ABC was dilated with the origin as the center of dilation to create triangle A'B'C'. The triangle was dilated using a scale factor of 14.


The coordinates of the vertices of triangle ABC are given. You can use the scale factor to find the coordinates of the dilated image.


Enter the coordinates of the vertices of triangle A'B'C' below.

(Decimal values may be used)
pls answer ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

Answer:

$A'(-0.25,0.75)\\\\B'(1.75,-0.25)\\\\C'(-1,-0.25)$

Step-by-step explanation:

A Dilation is defined as  a transformation in which the image and the pre-image have the same shape, but their sizes are different.

When the scale factor is greater than 1, the image obtained after the  dilation is greater than the pre-image and it is an "Enlargement".

When the scale factor is is between 0 and 1,  the image obtained after the  dilation is smaller than the pre-image and it is an "Reduction".

In this case you know that the triangle ABC was dilated by this scale factor:

$scale\ factor=\frac{1}{4}$

With the origin as the center of dilation.

Since:

$0<\frac{1}{4}<1$

It is a Reduction.

You can identify that the vertices of the triangle ABC are:

$A(-1,3)\\\\B(7,-1)\\\\C(-4,-1)$

So you need to multiply the coordinates of each vertex  of the triangle ABC by $\frac{1}{4}$, in order to get the coordinates of the triangle A'B'C. Then, you get:

$A'=(\frac{1}{4}(-1),\frac{1}{4}(3))=(-0.25,0.75)\\\\B'(\frac{1}{4}(7),\frac{1}{4}(-1))=(1.75,-0.25)\\\\C'(\frac{1}{4}(-4),\frac{1}{4}(-1))=(-1,-0.25)$