The triangle ABC, shown on the coordinate plane above, is dilated from the origin by scale factor r = 1/2. What is the location of triangle A'B'C'?

Answer:
see explanation
Step-by-step explanation:
Since the dilatation is centred at the origin with a scale factor of [tex]\frac{1}{2}[/tex].
Then the coordinates of the image are one half the coordinates of the original, that is
A(3, 4 ) → A'(1.5, 2 )
B(- 7, 2 ) → B'(- 3.5, 1 )
C(2, 2 ) → C'(1, 1 )
Scale factor is the number by which the a figure is smaller or bigger by the original shape of the figure. The [tex]\Delta(ABC)'[/tex] located at the point,
[tex]\left (2, \dfrac{3}{2} \right )\\[/tex],[tex]\left (\dfrac{-7}{2}, 1 \right )\\[/tex] , [tex](1,1)[/tex].
The [tex]\Delta ABC[/tex] is dilated by scale factor of 1/2.
The [tex]\Delta ABC[/tex] is located at the points A(2,2), B(4,3), and C(-7,2).
Scale factor is the number by which the a figure is smaller or bigger by the original shape of the figure.
As the point of the triangle shown in the figure is dilated with the scale factor of 1/2. Thus all the point will be multiplied by the scale factor 1/2,
Point A (4,3) after the dilation of scale factor of 1/2,
[tex]A'=\left (\dfrac{4}{2}, \dfrac{3}{2} \right )\\ A'=\left (2, \dfrac{3}{2} \right )\\[/tex]
Point B (-7,2) after the dilation of scale factor of 1/2,
[tex]B'=\left (\dfrac{-7}{2}, \dfrac{2}{2} \right )\\ B'=\left (\dfrac{-7}{2}, 1 \right )\\[/tex]
Point C (2,2) after the dilation of scale factor of 1/2,
[tex]C'=\left (\dfrac{2}{2}, \dfrac{2}{2} \right )\\ C'=(1,1)[/tex]
Hence the [tex]\Delta(ABC)'[/tex] located at the point,
[tex]\left (2, \dfrac{3}{2} \right )\\[/tex],[tex]\left (\dfrac{-7}{2}, 1 \right )\\[/tex] , [tex](1,1)[/tex]
Learn more about the scale factor here;
https://brainly.com/question/8765466