Find the affine transformation that maps the line (a)…

Find the affine transformation that maps the line $(a)\ x+y-1=0$ to $(a')\ x+y=0$, the line $(b)\ x-y=0$ to $(b')\ x-y-1=0$, and the point $A(0,0)$ to $A'(2,1)$.

Step 1: Translation part

Write $x'=a_{11}x+a_{12}y+b_1,\ y'=a_{21}x+a_{22}y+b_2$. The condition $A(0,0)\mapsto A'(2,1)$ gives $b_1=2,\ b_2=1$.

Step 2: Intersection point

The intersection $P=(a)\cap(b)=\left(\tfrac12,\tfrac12\right)$ must go to $P'=(a')\cap(b')=\left(\tfrac12,-\tfrac12\right)$: $\tfrac12(a_{11}+a_{12})+2=\tfrac12,\qquad \tfrac12(a_{21}+a_{22})+1=-\tfrac12$.

\[ a_{11}+a_{12}=-3,\qquad a_{21}+a_{22}=-3. \]

Now $A$ and $P$ both lie on $(b)$, and $A'$, $P'$ both lie on $(b')$. So the condition $(b)\to(b')$ is already satisfied.

Step 3: Line $(a)$

The points $R(1,0)$ and $S(0,1)$ of $(a)$ must go to points of $(a')$, i.e. $x'+y'=0$: \[(a_{11}+2)+(a_{21}+1)=0\ \Rightarrow\ a_{11}+a_{21}=-3,\qquad a_{12}+a_{22}=-3.\]

Together with step 2 this gives $a_{22}=a_{11}$ and $a_{12}=a_{21}=-3-a_{11}$. Of the four equations only three are independent, so one parameter $t=a_{11}$ remains free.

Step 4: General solution

\[x'=t\,x-(3+t)\,y+2,\qquad y'=-(3+t)\,x+t\,y+1,\qquad t\ne-\tfrac32\]

The determinant is $\Delta=t^2-(3+t)^2=-9-6t$, which must be non-zero, hence $t\ne-\tfrac32$.

Why a family? An affine map of the plane has 6 parameters. Here the conditions are not independent: $A$ lies on $(b)$, so $A\mapsto A'$ and $(b)\to(b')$ overlap. Only 5 independent conditions remain.

The simplest member is $t=0$, with $\Delta=-9$: $x'=-3y+2,\qquad y'=-3x+1$.

Step 5: Verification for $t=0$

• $A(0,0)\mapsto(2,1)=A'$.
• $R(1,0)\mapsto(2,-2)$ and $S(0,1)\mapsto(-1,1)$: both satisfy $x'+y'=0$.
• $(1,1)\mapsto(-1,-2)$ and $(2,2)\mapsto(-4,-5)$: both satisfy $x'-y'-1=0$.