Chapter 4.4, Problem 2CYP

To determine

The coordinates of $Z\text{'}$ using the given information.

Explanation of Solution

Given Information:

The coordinates of the vertices of the triangle XYZ is

  $\begin{array}{l}X(3,-1)\\ Y(-4,2)\\ Z(5,1)\end{array}$

The coordinates of the vertices of the triangle $X\text{'}Y\text{'}Z\text{'}$ is

  $\begin{array}{l}X\text{'}(1,0)\\ Y\text{'}(-6,3)\\ Z\text{'}(?,?)\end{array}$

Calculation:

Now let the coordinates of the vertex $Z\text{'}$ be,

  $Z\text{'}=(a,b)$

Now the vertex matrix of the triangle XYZ is,

  $\left|\begin{array}{ccc}3& -4& 5\\ -1& 2& 1\end{array}\right|$

Now the vertex matrix of the triangle $X\text{'}Y\text{'}Z\text{'}$ is,

  $\left|\begin{array}{ccc}1& -6& a\\ 0& 3& b\end{array}\right|$

Now let the transition matrix is,

  $\left|\begin{array}{ccc}x& x& x\\ y& y& y\end{array}\right|$

Now as per question,

  $\left|\begin{array}{ccc}3& -4& 5\\ -1& 2& 1\end{array}\right|+\left|\begin{array}{ccc}x& x& x\\ y& y& y\end{array}\right|=\left|\begin{array}{ccc}1& -6& a\\ 0& 3& b\end{array}\right|$

Now solving for x and y, we get,

  $\begin{array}{l}\left|\begin{array}{ccc}3+x& -4+x& 5+x\\ -1+y& 2+y& 1+y\end{array}\right|=\left|\begin{array}{ccc}1& -6& a\\ 0& 3& b\end{array}\right|\\ Now,\\ 3+x=1\\ or,x=1-3=-2\\ Also,\\ -1+y=0\\ or,y=1.\end{array}$

Now from the matrix using the values of x and y we get the values of a and b.

  $\begin{array}{l}5+x=a\\ 5-2=a\\ or,a=3.\\ Also,\\ 1+y=b\\ b=1+y\\ b=1+1=2.\end{array}$

Therefore the coordinates of $Z\text{'}$ is $(3,2)$