Graph the image of the figure after a dilation with a scale factor of 3 centered at (2,-7)
Graph the quadrilateral by connecting all its vertices.
(I'm not able to atach the image for some wierd reson so I will list the points of the orignal shape)
This is really important so if you could help it would be much appreicated!!
So you have a quad that looks like this. (Orange is original points, black is making up the shape)
Draw up 3 lines connecting the center to the 3 points. (I'm gonna use Red, Blue, and Green).
Since it's a dilation of 3, you're going to take each line and extend it 2 more times straight. Then mark the new points. (I'm going to use purple)
Connect the new lines.
The new points should be (2, -7), (-4, 3), (5, 5), and (8, -1).
Thanks so much for answering but you actually marked the 4th corner of the original shape wrong its expose to be (-2,-7) not (2,-7) however it is expose to still be centered at (2,-7). this should effect the answer do you think you could help me fix it so it is correct?
Well, we may not need that nice diagram to answer that question!
The fourth vertex of the quadrilateral is located at \((-2,-7)\), which is the same y-coordinate as the center of dilation, \((2,-7)\). Let's assume that the center of dilation is actually the origin of the coordinate plane.
This means that, from the original origin, we have effectively shifted the entire coordinate plane 2 units to the right and 7 units downward. Since the coordinates of the fourth vertex lie on the same y-coordinate, it is much easier to notice that the distance from the vertex to the "new origin" is 4 units. This means that the center of dilation is at \((0,0)\) and that the fourth vertex is located at \((-4,0)\)
If the question asks for a dilation by a scale factor of 3, then the distance of the vertex from our "new origin" also triples. The distance, therefore, would be 12, so the coordinate of the dilated point would be \((-12,0)\).
Of course, we must account for the fact that we shifted the Cartesian plane for convenience sake. In order to reverse these effects, we must move the plane 2 units to the left and 7 units upward, so the coordinates of the fourth vertex would become \((-10,-7)\)
That's quite a nice diagram, CoopTheDupe!
Unfortunately, my nitpicky mindset noticed that \((-2,-7)\), a coordinate of the original quadrilateral is graphed incorrectly (as \((2,7)\).
Here's a "formula" that can be used to find a new point :
New Point ' = [ Old Point - Dilation Center ] *Scale Factor + Dilation Center
Note that (a , b) - (c, d) = (a - c, b - d)
(a, b) + (c , d) = ( a + c, b + d) and
(a, b) * S = (Sa, Sb)
So.....to transform (0, - 4) as follows
New Point' = [ ( 0 - 4) - ( 2, - 7) ] * 3 + ( 2, -7) =
[ ( -2, 3) ] * 3 + (2, -7) =
(- 6, 9) + ( 2, - 7) =
You can compute the other points yourself