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Use matrices to determine the coordinates of the vertices of the reflected figure. Then graph the pre-image and the image on the same coordinate grid. (Picture provided)

Use matrices to determine the coordinates of the vertices of the reflected figure Then graph the preimage and the image on the same coordinate grid Picture prov class=
Use matrices to determine the coordinates of the vertices of the reflected figure Then graph the preimage and the image on the same coordinate grid Picture prov class=

Respuesta :

Answer:

The coordinates of the vertices of the reflected figure are :

R' is (5 , -2) , S' is (3 , 5) , T' is (-7 , 6) ⇒ the right answer is (d)

Step-by-step explanation:

* When you reflect a point across the line y = x, the x-coordinate

 and y-coordinate change their places.

- If the point is (x , y) then its image is (y , x)

* If you reflect over the line y = -x, the x-coordinate and y-coordinate

 change their places and their signs

- If the point is (x , y) then its image is (-y , -x)

* Lets study the matrix of the reflection  about the line y = x

- The matrix of the reflection about the line y = x is

 [tex]\left[\begin{array}{cc}0&1\\1&0\end{array}\right][/tex]

- Because the x-coordinate and y-coordinate change places.

* Now lets solve the problem

- We will multiply the matrix of the reflection about y = x

 by each point to find the image of each point

- The dimension of the matrix of the reflection about y = x

 is 2×2 and the dimension of the matrix of each point is 2×1,

 then the dimension of the matrix of each image is 2×1

∵ The point R is (-2 , 5)

∴ [tex]R'=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{cc}-2\\5\end{array}\right]=[/tex]

  [tex]\left[\begin{array}{c}(0)(-2)+(1)(5)\\(1)(-2)+(0)(5)\end{array}\right]=\left[\begin{array}{c}5\\-2\end{array}\right][/tex]

∴ R' is (5 , -2)

∵ The point S is (5 , 3)

∴ [tex]S'=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{c}5\\3\end{array}\right]=[/tex]

  [tex]\left[\begin{array}{c}(0)(5)+(1)(3)\\(1)(5)+(0)(3)\end{array}\right]=\left[\begin{array}{c}3\\5\end{array}\right][/tex]

∴ S' is (3 , 5)

∵ The point T is (6 , -7)

∴ [tex]T'=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{c}6\\-7\end{array}\right]=[/tex]

  [tex]\left[\begin{array}{c}(0)(6)+(1)(-7)\\(1)(6)+(0)(-7)\end{array}\right]=\left[\begin{array}{c}-7\\6\end{array}\right][/tex]

∴ T' is (-7 , 6)

* Lets look to the figures to find the right answer

∵ The R' is (5 , -2) , S' is (3 , 5) , T' is (-7 , 6)

∴ The right answer is (d)

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