Use scalar multiplication to determine the coordinates of the vertices of the dilated figure. Then graph the pre-image and the image of the same coordinate grid.
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Answer:
The coordinates of the vertices of the dilated figure are:
A' is (-2 , 4), B' is (4 , 8), C' is (4 , -2), D' is (-2 , -6) ⇒ the answer is (d)
Step-by-step explanation:
* Lets study the matrix of the dilation
- If we dilate any point by scale factor k we multiply the
coordinates of the point by k
- The matrix of the dilation by scale factor k is
[tex]\left[\begin{array}{ccc}k&0\\0&k\end{array}\right][/tex]
* Now lets solve the problem
- We will multiply the matrix of dilation by the matrix of the
vertices of the quadrilateral
- The dimension of the matrix of the dilation is 2×2 and the
dimension of the matrix of the vertices of the quadrilateral
is 2×4 then the dimension of the matrix of the image of the
quadrilateral is 2×4
∵ The scale factor is 2
∴ The matrix of dilation is [tex]\left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex]
∵ The matrix of the vertices of the quadrilateral is
[tex]\left[\begin{array}{cccc}-1&2&2&-1\\2&4&-1&-3\end{array}\right][/tex]
∴ The image of the quadrilateral is :
[tex]\left[\begin{array}{cc}2&0\\0&2\end{array}\right]\left[\begin{array}{cccc}-1&2&2&-1\\2&4&-1&-3\end{array}\right]=[/tex]
[tex]\left[\begin{array}{cccc}(2)(-1)+(0)(2)&(2)(2)+(0)(4)&(2)(2)+(0)(-1)&(2)(-1)+(0)(-3)\\(0)(-1)+(2)(2)&(0)(2)+(2)(4)&(0)(2)+(2)(-1)&(0)(-1)+(2)(-3)\end{array}\right]=[/tex]
[tex]\left[\begin{array}{cccc}-2&4&4&-2\\4&8&-2&-6\end{array}\right][/tex]
∴ The image of point A' is (-2 , 4)
∴ The image of point B' is (4 , 8)
∴ The image of point C' is (4 , -2)
∴ The image of point D' is (-2 , -6)
* The right answer is figure (d)
Answer:
d on edge 2021
Step-by-step explanation:
just took the test :)