Question

2.Original A ABC: A(1, 2), B(3, 4), C(3, 2) First image: A

Answer 1

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Answer:

  as written

• first image: (x, y) ⇒ (y, x+20)
• second image: (x, y) ⇒ (-10x+41, -y+24)
• overall: (x, y) ⇒ (-10y +41, -x+4)

  with B"(19,1)

• first image: (x, y) ⇒ (y, x+20)
• second image: (x, y) ⇒ (-x+23, -y+24)
• overall: (x, y) ⇒ (-y+23, -x+4)

Step-by-step explanation:

The transformation (x, y) ⇒ (ax+by+c, dx+ey+f) can be written as a square matrix ...

  $T=\left[\begin{array}{ccc}a&b&c\\d&e&f\\0&0&1\end{array}\right]$

operating on a triangle's coordinates ...

  $P=\left[\begin{array}{ccc}x_1&x_2&x_3\\y_1&y_2&y_3\\1&1&1\end{array}\right]$

That is, ...

  P' = T·P

Post-multiplying by P⁻¹ gives the transformation matrix:

  T = P'·P⁻¹

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For the transformation from the original image P to the first image P', we have ...

  $T_1\cdot\left[\begin{array}{ccc}1&3&3\\2&4&2\\1&1&1\end{array}\right] =\left[\begin{array}{ccc}2&4&2\\21&23&23\\1&1&1\end{array}\right] \ \Rightarrow\ T_1=\left[\begin{array}{ccc}0&1&0\\1&0&20\\0&0&1\end{array}\right]$

This represents a reflection across y=x followed by translation upward 20 units.

The transformation from the first image P' to the second image P" is then ...

  $T_2\cdot\left[\begin{array}{ccc}2&4&2\\21&23&23\\1&1&1\end{array}\right] =\left[\begin{array}{ccc}21&1&21\\3&1&1\\1&1&1\end{array}\right] \ \Rightarrow\ T_2=\left[\begin{array}{ccc}-10&0&41\\0&-1&24\\0&0&1\end{array}\right]$

This represents a reflection across the origin, a horizontal stretch by a factor of 10, then a translation right 41 and up 24.

The original, first image, and second image are plotted in red, green, and blue in the first attachment,

Perhaps more conventionally described, the transformations are ...

• first image: (x, y) ⇒ (y, x+20)
• second image: (x, y) ⇒ (-10x+41, -y+24)
• overall: (x, y) ⇒ (-10y +41, -x+4)

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If the coordinates of B" represent a typo, and if the intended coordinates are B"(19, 1), then the second transformation is ...

  $T_{2a}\cdot\left[\begin{array}{ccc}2&4&2\\21&23&23\\1&1&1\end{array}\right] =\left[\begin{array}{ccc}21&19&21\\3&1&1\\1&1&1\end{array}\right] \ \Rightarrow\ T_{2a}=\left[\begin{array}{ccc}-1&0&23\\0&-1&24\\0&0&1\end{array}\right]$

This, too, is a reflection across the origin, followed by a translation right 23 and up 24. There is no stretching involved here. The conventional description of the transformations might be ...

• first image: (x, y) ⇒ (y, x+20)
• second image: (x, y) ⇒ (-x+23, -y+24)
• overall: (x, y) ⇒ (-y+23, -x+4)

The images are shown in the second attachment.

_____

Additional comment

The equations for the transformations can be solved by hand, but it is more convenient to use a machine solver of some sort. Many spreadsheet programs have the ability to create an inverse matrix and to multiply matrices. Many graphing calculators can do this, too. On-line solvers are available for the same purpose.