Chapter 12.6, Problem 26E

To determine

To find:the coordinates of the vertices of the original triangle.

Answer to Problem 26E

the coordinate of original triangle,

  $\begin{array}{l}\left(x,y\right)\to \left(-y,x\right)\\ A\text{'}\left(-2,-3\right)\to A\text{'}\text{'}\left(3,-2\right)\\ B\text{'}\left(0,-1\right)\to B\text{'}\text{'}\left(1,0\right)\\ C\text{'}\left(3,-4\right)\to C\text{'}\text{'}\left(4,3\right)\end{array}$

Explanation of Solution

Given : theverticesprovided in the question,

Final image vertices = $\left(1,-4\right),\left(3,-2\right),\left(6,-5\right)$

Using translator = $\left(x,y\right)\to \left(x+3,y-1\right)$

Calculation : first using translator from final image coordinates vertices, we find the rotated image vertices,

  $\left(x,y\right)\to \left(x+3,y-1\right)$

  $\begin{array}{l}\left(x,y\right)\to \left(x-3,y+1\right)\\ A\left(1,-4\right)\to A\text{'}\left(-2,-3\right)\\ B\left(3,-2\right)\to B\text{'}\left(0,-1\right)\\ C\left(6,-5\right)\to C\text{'}\left(3,-4\right)\end{array}$

Now for the original image, switch the coordinates and x -coordinate of image multiply by -1.

  $\begin{array}{l}\left(x,y\right)\to \left(-y,x\right)\\ A\text{'}\left(-2,-3\right)\to A\text{'}\text{'}\left(3,-2\right)\\ B\text{'}\left(0,-1\right)\to B\text{'}\text{'}\left(1,0\right)\\ C\text{'}\left(3,-4\right)\to C\text{'}\text{'}\left(4,3\right)\end{array}$

Hence,

the coordinate of original triangle,

  $\begin{array}{l}\left(x,y\right)\to \left(-y,x\right)\\ A\text{'}\left(-2,-3\right)\to A\text{'}\text{'}\left(3,-2\right)\\ B\text{'}\left(0,-1\right)\to B\text{'}\text{'}\left(1,0\right)\\ C\text{'}\left(3,-4\right)\to C\text{'}\text{'}\left(4,3\right)\end{array}$