What transformation maps $y=\frac{1}{4}x-2$ to $y=-3x+6$? What transformation maps $y=\frac{1}{4}x-2$ to $y=-3x+6$?
I have tried many things, rotating around the origin, reflecting about common lines, and nothing seems to work.
Any help is appreciated. Thanks!
 A: HINT 
The transformation will be a rotation about the intersection of $\frac14 x -2$ and $-3x+6$.
A: Have a look at the geometrical situation:

(Large version)
We have 
$f(x) =\frac{1}{4}x-2$ (green line) and $g(x) =-3x+6$ (red line).
We see that a rotation around the intersection $A$ of both lines would do most of the job.
Easiest, rotations around the origin are formulated, so we 


*

*first translate the scene such that $A$ turns into the origin $O$ via $T$, then 

*rotate the line of $f$ into the line of $g$ via $R$ around the origin, and finally 

*retranslate the origin $O$ to $A$ again via $T^{-1}$.


These individual transformations compose into the transformation
$$
M = T^{-1} R T
$$
In Detail:
As we want to use translations, in other words: affine transformations, and still use the convenient matrix formulation, we use homogeneous coordinates $u = (x, y, 1)^\top$. 
Then the transformation matrices are:
$$
T =
\begin{pmatrix}
1 & 0 & -a_x \\
0 & 1 & -a_y \\
0 & 0 & 1
\end{pmatrix}
$$
where $A = (a_x, a_y)$. Test:
$$
T u = 
\begin{pmatrix}
1 & 0 & -a_x \\
0 & 1 & -a_y \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
1
\end{pmatrix}
=
\begin{pmatrix}
x-a_x \\
y-a_y \\
1
\end{pmatrix}
$$
so $T(u_A) = (0,0,1)^\top$, as desired. The rotation around the origin about an angle $\varphi$ is
$$
R = 
\begin{pmatrix}
\cos(\varphi) & -\sin(\varphi) & 0 \\
\sin(\varphi) & \cos(\varphi) & 0 \\
0 & 0 & 1
\end{pmatrix}
$$
The inverse transformation of $T$ is
$$
T^{-1} =
\begin{pmatrix}
1 & 0 & a_x \\
0 & 1 & a_y \\
0 & 0 & 1
\end{pmatrix}
$$
Calculation of the parameters:
We can describe the two lines by
\begin{align}
(1/4) x - y &= 2 \\
3x + y &= 6
\end{align}
which we can solve as
$$
\left[
\begin{array}{rr|r}
1/4 & -1 & 2 \\
3 & 1 & 6
\end{array}
\right]
\to
\left[
\begin{array}{rr|r}
1 & -4 & 8 \\
3 & 1 & 6
\end{array}
\right]
\to
\left[
\begin{array}{rr|r}
1 & -4 & 8 \\
0 & 13 & -18
\end{array}
\right]
\to
\left[
\begin{array}{rr|r}
1 & -4 & 8 \\
0 & 1 & -18/13
\end{array}
\right]
\to
\left[
\begin{array}{rr|r}
1 & 0 & 32/13 \\
0 & 1 & -18/13
\end{array}
\right]
$$
thus $A = (32/13,-18/13) = (2.4615,-1.3846)$.
We can determine the angle between the lines from
$$
d_f \cdot d_g = \cos(\angle(d_f, d_g)) = \cos(-\varphi)
$$ 
and the unit direction vectors of the lines, which are
$$
d_f 
= \frac{(1, 1/4)^\top}{\lVert (1, 1/4) \rVert} 
= \frac{(4,1)^\top}{\sqrt{17}}
$$ 
and 
$$
d_g 
= \frac{(1,-3)^\top}{\lVert (1,-3) \rVert}
= \frac{(1,-3)^\top}{\sqrt{10}}
$$
so $\varphi = -\arccos(1/\sqrt{170})=-85.6^\circ$.
A: A single $2\times2$ matrix transformation will do the job.
Consider mapping $\binom 80$ to $\binom 06$ and $\binom 0{-2}$ to $ \binom 20$
In which case the required matrix  $M$ is given by $$M\left(\begin{matrix}8&0\\0&-2\end{matrix}\right)=\left(\begin{matrix}0&2\\6&0\end{matrix}\right)$$
Post-multiplying by the inverse and so on leads to $$M=\left(\begin{matrix}0&-1\\ \frac 34&0\end{matrix}\right)$$
A: If we extend (but also limit) our research to affine transformations i.e., transformations $(x,y)  \longrightarrow (x',y')$ such that:
$$\tag{1}\begin{cases}x'&=&ax+cy+e\\y'&=&bx+dy+f\end{cases}$$
for certain fixed coefficients $a,b,c,d,e,f$.
Such transformations are known to map lines onto lines.
Thus, it suffices to take 


*

*two points $A_1(0,-2), B_1(8,0)$ on the first line $(L_1)$ with equation $y=\frac{1}{4}x-2.$

*two points $A_2(0,6), B_2(2,0)$ on the second line $(L_2)$ with equation $y=-3x+6.$
Due to (1), imposing that the image of $A_1$ and $B_1$ is $A_2$ and $B_2$ resp.  gives 4 equations in the 6 unknowns $a,b,c,d,e,f,g$. 
$$\tag{2}\begin{cases}0&=&-2c+e\\6&=&-2d+f\\2&=&8a+e\\0&=&8b+f\end{cases}$$
Thus there is an infinite number of solutions, that can be expressed as function of arbitrary coefficients $a$ and $b$ for example, giving:
$$\tag{3}\begin{cases}x'&=&ax+(-4a+1)y+(-8a+2)\\y'&=&bx+(-4b-3)y+(-8b)\end{cases}$$
Remarks : there would be many remarks to be done about the choices that have been made.
1) As we have not realized an isometric mapping (distance $A_1B_1$ is not equal to distance $A_2B_2$), one should not be surprized not to be able to find in (3) any $\pm\frac{\pi}{2}$ rotation.
2) There is a large degree of arbitraryness in the choice of points on lines $(L_1)$ and $(L_2)$. One can say that the issue depends on four more degrees of freedom that correspond e.g., to arbitrary choices for abscissas of points $A_1, B_1, A_2, B_2$.
3) The most general transformations that map lines onto lines are projective transformations :
$$\tag{4}\begin{cases}x'&=&\frac{ax+cy+e}{gx+hy+k}\\y'&=&\frac{bx+dy+f}{gx+hy+k}\end{cases}$$
giving 3 (in fact only two degrees of freedom).
