What transformation maps $y=\frac{1}{4}x-2$ to $y=-3x+6$?

Question

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!

Answer 1

HINT

The transformation will be a rotation about the intersection of $\frac14 x -2$ and $-3x+6$.

Answer 2

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 $\varphi = -\arccos(1/\sqrt{170})=-85.6^\circ$.

Answer 3

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 $$M=\left(\begin{matrix}0&-1\\ \frac 34&0\end{matrix}\right)$$

Answer 4

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).