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.
On
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 $(1-t)L+tK$.
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 $f(3,4)$.