Answer:
(-15,12),(6,18)(12,-9)
Step-by-step explanation:
Given
A (-5, -4), B (2, 6), and C (4, -3)
Dilated from (x,y) to (3x,3y)
Required
Vertices of the dilated image
First, the scale factor has to be calculated
This is calculated as thus;
Dilation = Original * Scale Factor;
In case of (x,y) to (3x,3y)
So, we have
[tex](3x,3y) = Scale\ Factor * (x,y)[/tex]
Simplify brackets
[tex]3(x,y) = Scale\ Factor * (x,y)[/tex]
Divide both sides by (x,y)
[tex]\frac{3(x,y)}{(x,y)} = \frac{Scale\ Factor * (x,y)}{(x,y)}[/tex]
[tex]\frac{Scale\ Factor * (x,y)}{(x,y)}=\frac{3(x,y)}{(x,y)}[/tex]
[tex]Scale\ Factor =\frac{3(x,y)}{(x,y)}[/tex]
[tex]Scale\ Factor =3[/tex]
Now, the vertices of the new image can be calculated;
Using the formula: Dilation = Original * Scale Factor;
At vertex A(-5,4)
Dilation = (-5,4) * 3
Dilation = (-5*3 ,4*3)
Dilation = (-15,12)
At vertex B(2,6)
Dilation = (2,6) * 3
Dilation = (2*3 ,6*3)
Dilation = (6,18)
At vertex C(4,-3)
Dilation = (4,-3) * 3
Dilation = (4*3 ,-3*3)
Dilation = (12,-9)
Hence, the vertices of the dilated image are (-15,12),(6,18)(12,-9)
