Corresponding Point for a Glide Reflection I was wondering if there was an efficient method that could solve these types of problems. 
Here is the problem:
Plot the points K = (0,0), L = (7,-1), M = (9,3), P = (6,7), Q = (10,5), and R = (1,2). You will see that the triangles KLM and RPQ are congruent. Find coordinates for the point in triangle KLM that corresponds to (3,4) in triangle RPQ.
 A: What helps is to know the general form for an isometry of the Euclidean plane:
\begin{align*}f(x,y) &= (x,y) \begin{pmatrix} a & b \\ c & d \end{pmatrix} + (e,f) \\
  &= (ax+cy+e,bx+dy+f)
\end{align*}
such that the matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is is orthonormal, meaning that $a^2+b^2=1$, $c^2+d^2=1$ and $ac+bd=0$. In addition to these three equations, you get three more equations $f(K)=R$, $f(L)=P$, $f(M)=Q$ by plugging in the coordinates of $K,L,M,R,P,Q$. You can then solve for the six unknowns $a,b,c,d,e,f$. Substituting those back into the formula for $f(x,y)$, you can then compute $f(3,4)$. 
A: Working in homogeneous coordinates, the image of $(3,4)$ under the affine map that sends $\triangle{RPQ}$ to $\triangle{LMK}$ is (writing all of the coordinates as column vectors) $$\begin{bmatrix}K&L&M \\ 1&1&1 \end{bmatrix} \begin{bmatrix}R&P&Q\\1&1&1\end{bmatrix}^{-1}\begin{bmatrix}3\\4\\1\end{bmatrix}.$$ However, there’s a much simpler solution to this particular problem.  
Once you’ve plotted these points as instructed, observe that $(3,4)$ lies on the segment $\overline{PR}$ (you can easily verify this algebraically). Since $\overline{PR}\cong\overline{LK}$, you just have to find the point on the latter segment that’s the same distance from $L$ as $(3,4)$ is from $P$. Equivalently, solve for $t$ in $(1-t)P+tR=(3,4)$ and plug that value into $(1-t)L+tK$.
