Find the image vertices for a dilation with center (0,0)
and a scale factor of 4.

* show steps if you can. if not, it’s fine

Image vertices are : A' = (-12 , 4) ; B' = (16 , -12) ; C' = (8 , 12) ; D' = (-4 , 16)

Step-by-step explanation:

Given :

Scale Factor = 4

Vertices given in the above graph :

A = (-3 , 1) ; B = (4 , -3) ; C = (2 , 3) ; D = (-1 , 4)

Calculation :

To find the image vertices for a dilation with center (0 , 0) we have to multiply the vertices by scale factor 4.

So the image vertices are :

A' = (-12 , 4) ; B' = (16 , -12) ; C' = (8 , 12) ; D' = (-4 , 16)

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The resultant image vertices will be,

A' = (-12 ,4) ;  B' = (16 , -12) ; C' = (8, 12) and D'= (-4, 16)

Step-by-step explanation:

The co-ordinates of the given image vertices are,

A = (-3 , 1) ; B = (4 ,-3); C = (2, 3) and D = (-1, 4)

If the image is dilated with center (0, 0) and a scale factor of 4, then in the resultant image, all the co-ordinates will be multiplied by a factor of 4.

So, the resultant image vertices will be,

A' = (-12 ,4) ;  B' = (16 , -12) ; C' = (8, 12) and D'= (-4, 16)

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

y=3x+4

step-by-step explanation:

given the points a(0,4) and b(-4, -8)

eqn of line ab: (y-4)/(x-0)=(-8-4)/(-4-0)

(y-4)/x=-12/-4=3

y=3x+4