Introduction to Projective Geometry Solutions 5.2

Please read this introduction first before looking through the solutions. Here’s a quick index to all the problems in this section.

1 2 3 4 5 6 7 8 9 10

1. Verify that under a transformation of point-coordinates the equation of a line always goes into a linear equation in the new coordinates. Is the corresponding result true for the equation of a point?

Consider the equation of a general line $\Lambda^TX = 0$ where $\Lambda$ is the column vector of the line coordinates and $X$ is the column vector of the point coordinates. Suppose the point-coordinates were transformed to $X’$ by the matrix of transformation $A$. Then, $$X’ = AX$$ and, $$X = A^{-1}X’$$ Substituting this in the equation of the line we get $$\Lambda^TA^{-1}X’ = 0$$ Now $\Lambda^TA^{-1}$ is row vector that is a linear combination of the rows of $A^{-1}$. So writing $(\Lambda^TA^{-1})^T$ as $\Lambda’$ we get, $$\Lambda’^TX’ = 0$$ It is evident from this form that the equation of the line in the new point-coordinates is still linear in the new point-coordinates $X’$.

Suppose the line-coordinates were transformed to $\Lambda’$ by the matrix of transformation $B$. Then, $$\Lambda’ = B\Lambda$$ and, $$\Lambda = B^{-1}\Lambda’$$ Substituting this in the equation of the line we get $$(B^{-1}\Lambda’)^TX = 0$$ $$\implies \Lambda’^T(B^{-1})^TX = 0$$ Now $(B^{-1})^TX$ is column vector that is a linear combination of the columns of $(B^{-1})^T$. So writing $(B^{-1})^TX$ as $X’$ we get, $$\Lambda’^TX’ = 0$$ It is evident from this form that the equation of the line in the new line-coordinates is still linear in the new line-coordinates $\Lambda’$.

2. Verify that the form of the condition that three points be collinear is preserved under any transformation of point-coordinates.

Let the three points be $X_1$, $X_2$, $X_3$. Let’s represent the general transformation matrix by $A$. So we have the new coordinates as $$X’_1 = AX_1, X_1 = A^{-1}X’_1$$ $$X’_2 = AX_2, X_2 = A^{-1}X’_2$$ $$X’_3 = AX_3, X_3 = A^{-1}X’_3$$

If these three points are collinear then by Theorem 1, Corollary 1, Sec. 4.2 we have

$$\begin{vmatrix} X_1 & X_2 & X_3 \end{vmatrix} = 0$$ $$\implies \begin{vmatrix} A^{-1}X’_1 & A^{-1}X’_2 & A^{-1}X’_2\end{vmatrix} = 0$$ $$\implies \begin{vmatrix} A^{-1} \end{vmatrix} \begin{vmatrix} X’_1 & X’_2 & X’_2 \end{vmatrix} = 0$$ As $A$ is by definition a non-singular matrix, its (or its inverse’s) determinant can not be zero. Hence $$\begin{vmatrix} X’_1 & X’_2 & X’_2 \end{vmatrix} = 0$$

This is same form as prescribed by Theorem 1, Corollary 1, Sec 4.2. Hence the form of the condition that three points be collinear is preserved under any transformation of point-coordinates.

3. Verify that the form of the condition that three lines be concurrent is preserved under any transformation of line-coordinates.

Easy! Just replace the words “point” and “collinear” with the words “line” and “concurrent” and use Theorem 2, Corollary 1, Sec 4.2 instead of Theorem 1 in the solution to the previous exercise and you can verify the given statement.

4. If $\lambda_1: x_1 + x_2 + x_3 = 0$, $\lambda_2: x_2 + x_3 = 0$, and $\lambda_3: x_1 - x_3 = 0$ are chosen as the $x’_1$, $x’_2$, and $x’_3$ axes, respectively, and if $U:(1, -1, 2)$ is chosen as the new unit point, find the equations expressing the new point-coordinates in terms of the old, and vice versa.

The equations expressing the new point-coordinates in terms of the old will be

$$x’_1 = h_1(x_1 + x_2 + x_3)$$ $$x’_2 = h_2(x_2 + x_3)$$ $$x’_3 = h_3(x_1 - x_3)$$

provided $h_1, h_2, h_3$ are determined so that the new coordinates of the point $(1, -1, 2)$ are $(1, 1, 1)$.

That means $$1 = h_1(1 - 1 + 2) \implies h_1 = \frac{1}{2}$$ $$1 = h_2(-1 + 2) \implies h_2 = 1$$ $$1 = h_3(1 - 2) \implies h_3 = -1 $$

Hence the equation of the point coordinate transformation is

$$X’ = AX$$ $$A = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ -1 & 0 & 1 \end{pmatrix}$$ $$A^{-1} = \begin{pmatrix} 2 & -1 & 0 \\ -2 & 2 & -1 \\ 2 & -1 & 1 \end{pmatrix}$$

5. Given the euclidean circles $C_1: x^2 + y^2 = \frac{1}{2}$, $C_2: x^2 + y^2 = 1$, and $C_3: x^2 + y^2 = 4$. Find the equations of each of these circles if the coordinate transformation of Exercise 4 is specialized so that the ideal line is:
(a) $\lambda_1$ (b) $\lambda_2$ (c) $\lambda_3$.
What type of conic does each of the new equations represent? Why?

Let’s represent these equations in homogeneous coordinates with $x = \frac{x_1}{x_3}$ and $y = \frac{x_2}{x_3}$.

$$C_1: 2x_1^2 + 2x_2^2 - x_3^2 = 0$$ $$C_2: x_1^2 + x_2^2 - x_3^2 = 0$$ $$C_3: x_1^2 + x_2^2 - 4x_3^2 = 0$$

From the solution to the previous exercise, we have the following equations of transformation. $$x_1 = 2x’_1 - x’_2$$ $$x_2 = -2x’_1 + 2x’_2 - x’_3$$ $$x_3 = 2x’_1 - x’_2 + x’_3$$

Expressing this in the new point-coordinate system we get $$C_1: 2(2x’_1 - x’_2)^2 + 2(-2x’_1 + 2x’_2 - x’_3)^2 - (2x’_1 - x’_2 + x’_3)^2 = 0$$ $$C_2: (2x’_1 - x’_2)^2 + (-2x’_1 + 2x’_2 - x’_3)^2 - (2x’_1 - x’_2 + x’_3)^2 = 0$$ $$C_3: (2x’_1 - x’_2)^2 + (-2x’_1 + 2x’_2 - x’_3)^2 - 4(2x’_1 - x’_2 + x’_3)^2 = 0$$

(a) If $\lambda_1$ is the ideal line then $x’_1 \ne 0$ in this specialization and we can divide by $x’_1$ to obtain the non-homogeneous representation. Representing $x = \frac{x’_2}{x’_1}$ and $y = \frac{x’_3}{x’_1}$ and dividing by $(x’_1)^2$ we get

$$C_1: 2(2 - x)^2 + 2(-2 + 2x - y)^2 - (2 - x + y)^2 = 0$$ $$C_2: (2 - x)^2 + (-2 + 2x - y)^2 - (2 - x + y)^2 = 0$$ $$C_3: (2 - x)^2 + (-2 + 2x - y)^2 - 4(2 - x + y)^2 = 0$$

Expanding, we get $$C_1: 9x^2 - 6xy + y^2 - 20x + 4y + 12 = 0 $$ $$C_2: 4x^2 - 2xy - 8x + 4 = 0 $$ $$C_3: x^2 + 4xy - 3y^2 + 4x - 12y - 8 = 0 $$

Hence, based on their discriminants (0, 132, 28), $C_1$ is a parabola and $C_2$ and $C_3$ are hyperbolas. Note that $C_2$’s equation is wrong in the book for this case. The coefficient of the $xy$ term should be $-2$ instead of $1$.

(b) If $\lambda_2$ is the ideal line then $x’_2 \ne 0$ in this specialization and we can divide by $x’_2$ to obtain the non-homogeneous representation. Representing $x = \frac{x’_1}{x’_2}$ and $y = \frac{x’_3}{x’_2}$ and dividing by $(x’_2)^2$ we get

$$C_1: 2(2x - 1)^2 + 2(-2x + 2 - y)^2 - (2x - 1 + y)^2 = 0$$ $$C_2: (2x - 1)^2 + (-2x + 2 - y)^2 - (2x - 1 + y)^2 = 0$$ $$C_3: (2x - 1)^2 + (-2x + 2 - y)^2 - 4(2x - 1 + y)^2 = 0$$

Expanding, we get $$C_1: 12x^2 + 4xy + y^2 - 20x - 6y + 9 = 0 $$ $$C_2: 4x^2 - 8x - 2y + 4 = 0$$ $$C_3: 8x^2 + 12xy + 3y^2 - 4x - 4y - 1 = 0 $$

Hence, based on their discriminants (-32, 0, 48), $C_1$ is an ellipse, $C_2$ is a parabola and $C_3$ is a hyperbola. Note that $C_2$’s equation is wrong in the book for this case too. The coefficient of the $y$ term should be $-2$ instead of $1$.

(c) If $\lambda_3$ is the ideal line then $x’_3 \ne 0$ in this specialization and we can divide by $x’_3$ to obtain the non-homogeneous representation. Representing $x = \frac{x’_1}{x’_3}$ and $y = \frac{x’_2}{x’_3}$ and dividing by $(x’_3)^2$ we get

$$C_1: 2(2x - y)^2 + 2(-2x + 2y - 1)^2 - (2x - y + 1)^2 = 0$$ $$C_2: (2x - y)^2 + (-2x + 2y - 1)^2 - (2x - y + 1)^2 = 0$$ $$C_3: (2x - y)^2 + (-2x + 2y - 1)^2 - 4(2x - y + 1)^2 = 0$$

Expanding, we get $$C_1: 12x^2 - 20xy + 9y^2 + 4x - 6y + 1 = 0 $$ $$C_2: 4x^2 - 8xy + 4y^2 - 2y = 0 $$ $$C_3: 8x^2 - 4xy - y^2 + 12x - 4y + 3 = 0 $$

Hence, based on their discriminants (-32, 0, 48), $C_1$ is an ellipse, $C_2$ is a parabola and $C_3$ is a hyperbola. Note that $C_2$’s equation is wrong in the book for this case too. The coefficient of the $y$ term should be $-2$ instead of $1$.

6. Given $\lambda_1: x_3 = 0$, Given $\lambda_2: x_1 - x_2 = 0$, Given $\lambda_3: x_1 - x_3 = 0$, $U:(0, 1, 1)$, and $v:[2, -1, -2]$. (a) Find the equations of the point-coordinate transformation for which $\lambda_1$, $\lambda_2$, and $\lambda_3$ are, respectively, the $x’_1$, $x’_2$, $x’_3$ axes and for which $U$ is the new unit point. Find the equations of the line-coordinate transformation for which $\lambda_2\lambda_3$, $\lambda_3\lambda_1$, $\lambda_1\lambda_2$ are respectively, the points $l’_1 = 0$, $l’_2 = 0$, $l’_3 = 0$, and for which $v$ is the new unit line. What is the incidence condition in the new coordinates?
(b) Work part (a) if $\lambda_2\lambda_3$, $\lambda_1\lambda_2$, $\lambda_1\lambda_3$ are respectively, the points $l’_1 = 0$, $l’_2 = 0$, $l’_3 = 0$.
(c) Work part (a) if $\lambda_1\lambda_3$, $\lambda_1\lambda_2$, $\lambda_2\lambda_3$ are respectively, the points $l’_1 = 0$, $l’_2 = 0$, $l’_3 = 0$

The equations expressing the new point-coordinates in terms of the old will be

$$x’_1 = h_1(x_3)$$ $$x’_2 = h_2(x_1 - x_2)$$ $$x’_3 = h_3(x_1 - x_3)$$

provided $h_1, h_2, h_3$ are determined so that the new coordinates of the point $(0, 1, 1)$ are $(1, 1, 1)$.

That means $$1 = h_1(1) \implies h_1 = 1$$ $$1 = h_2(0 - 1) \implies h_2 = -1$$ $$1 = h_3(0 - 1) \implies h_3 = -1 $$

Hence the equation of the point coordinate transformation is

$$X’ = AX$$ $$A = \begin{pmatrix} 0 & 0 & 1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix}$$ $$A^{-1} = \begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & -1 \\ 1 & 0 & 0 \end{pmatrix}$$

For the transformation of line-coordinates, let’s get the coordinates of the lines that are to be vertices in the new coordinate system.

$$\lambda_2\lambda_3: (1, -1, 0) \times (1, 0, -1) = [1, 1, 1]$$ $$\lambda_1\lambda_3: (0, 0, 1) \times (1, 0, -1) = [0, 1, 0]$$ $$\lambda_1\lambda_2: (0, 0, 1) \times (1, -1, 0) = [1, 1, 0]$$

Hence the vertices of the new coordinate system will be, $$l_1 + l_2 + l_3 = 0 $$ $$l_2 = 0 $$ $$l_1 + l_2 = 0 $$

The equations expressing the new line-coordinates in terms of the old will be

$$l’_1 = k_1(l_1 + l_2 + l_3)$$ $$l’_2 = k_2(l_2)$$ $$l’_3 = k_3(l_1 + l_2)$$

provided $k_1, k_2, k_3$ are determined so that the new coordinates of the line $[2, -1, -2]$ are $[1, 1, 1]$.

That means $$1 = k_1(2 - 1 - 2) \implies k_1 = -1 $$ $$1 = k_2(-1) \implies k_2 = -1 $$ $$1 = k_3(2 - 1) \implies k_3 = 1 $$

Hence the equation of the line coordinate transformation is

$$\Lambda’ = B\Lambda$$ $$B = \begin{pmatrix} -1 & -1 & -1 \\ 0 & -1 & 0 \\ 1 & 1 & 0 \end{pmatrix}$$ $$B^{-1} = \begin{pmatrix} 0 & 1 & 1 \\ 0 & -1 & 0 \\ -1 & 0 & -1 \end{pmatrix}$$

The incidence condition in the old coordinate system is

$$\Lambda^TX = 0$$ $$\implies (B^{-1}\Lambda’)^T(A^{-1}X’) = 0$$ $$\implies \Lambda’^T(B^{-1})^TA^{-1}X’ = 0$$

Now $(B^{-1})^TA^{-1}$ = $$ \begin{pmatrix} 0 & 1 & 1 \\ 0 & -1 & 0 \\ -1 & 0 & -1 \end{pmatrix}^T \begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & -1 \\ 1 & 0 & 0 \end{pmatrix} $$ $$ = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$

So the incidence condition will be $$ l’_1x’_1 + l’_2x’_2 + l’_3x’_3 = 0$$

Hence the form of the incidence condition is preserved.

(b) The equations expressing the new line-coordinates in terms of the old will be

$$l’_1 = k_1(l_1 + l_2 + l_3)$$ $$l’_2 = k_2(l_1 + l_2)$$ $$l’_3 = k_3(l_2)$$

provided $k_1, k_2, k_3$ are determined so that the new coordinates of the line $[2, -1, -2]$ are $[1, 1, 1]$.

That means $$1 = k_1(2 - 1 - 2) \implies k_1 = -1 $$ $$1 = k_2(2 - 1) \implies k_2 = 1 $$ $$1 = k_3(-1) \implies k_3 = -1 $$

Hence the equation of the line coordinate transformation is

$$\Lambda’ = B\Lambda$$ $$B = \begin{pmatrix} -1 & -1 & -1 \\ 1 & 1 & 0 \\ 0 & -1 & 0 \end{pmatrix}$$ $$B^{-1} = \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & -1 \\ -1 & -1 & 0 \end{pmatrix}$$

$(B^{-1})^TA^{-1}$ = $$ \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & -1 \\ -1 & -1 & 0 \end{pmatrix}^T \begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & -1 \\ 1 & 0 & 0 \end{pmatrix} $$ $$ = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix} $$

So the incidence condition will be $$ l’_1x’_1 + l’_2x’_3 + l’_3x’_2 = 0$$

(c) The equations expressing the new line-coordinates in terms of the old will be

$$l’_1 = k_1(l_2)$$ $$l’_2 = k_2(l_1 + l_2)$$ $$l’_3 = k_3(l_1 + l_2 + l_3)$$

provided $k_1, k_2, k_3$ are determined so that the new coordinates of the line $[2, -1, -2]$ are $[1, 1, 1]$.

That means $$1 = k_1(-1) \implies k_1 = -1 $$ $$1 = k_2(2 - 1) \implies k_2 = 1 $$ $$1 = k_3(2 - 1 - 2) \implies k_3 = -1 $$

Hence the equation of the line coordinate transformation is

$$\Lambda’ = B\Lambda$$ $$B = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & -1 \end{pmatrix}$$ $$B^{-1} = \begin{pmatrix} 1 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & -1 & -1 \end{pmatrix}$$

$(B^{-1})^TA^{-1}$ = $$ \begin{pmatrix} 1 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & -1 & -1 \end{pmatrix}^T \begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & -1 \\ 1 & 0 & 0 \end{pmatrix} $$ $$ = \begin{pmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ -1 & 0 & 0 \end{pmatrix} $$

So the incidence condition will be $$ l’_1x’_2 + l’_2x’_3 + l’_3x’_1 = 0$$

7. (a) Find the equations of the coordinate transformation which assigns to the points $P_1:(1, 1, -1)$, $P_2:(1, 0, 2)$ $P_1:(1, -1, 0)$, $U:(2, -1, 1)$ the new coordinates $(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1),$ respectively.
(b) Work part (a), given $P_1:(1, 0, -1)$, $P_2:(0, 1, 1)$, $P_3:(1, 1, 1)$, $U:(1, 2, 0)$.

We know from the solution to #1 that the equation of a line is transformed into a linear equation under a transformation of point-coordinates. Hence, from the transformations given, it is clear that the lines $P_1P_2$, $P_2P_3$ and $P_3P_1$ are the axes of the new coordinate system.

Computing these line coordinates, we get

$$ P_1P_2: (1, 1, -1) \times (1, 0, 2) = [2, -3, -1] $$ $$ P_2P_3: (1, 0, 2) \times (1, -1, 0) = [2, 2, -1] $$ $$ P_3P_1: (1, -1, 0) \times (1, 1, -1) = [1, 1, 2] $$

As $P’_1P’_2$, $P’_2P’_3$, $P’_3P’_1$ are [0, 0, 1], [1, 0, 0] and [0, 1, 0] respectively, $P_1P_2$, $P_2P_3$, $P_3P_1$ will be the axes $x’_3$, $x’_1$, $x’_2$ respectively.

The equations expressing the new point-coordinates in terms of the old will be

$$x’_1 = h_1(2x_1 + 2x_2 - x_3)$$ $$x’_2 = h_2(x_1 + x_2 + 2x_3)$$ $$x’_3 = h_3(2x_1 - 3x_2 - x_3)$$

provided $h_1, h_2, h_3$ are determined so that the new coordinates of the point $(2, -1, 1)$ are $(1, 1, 1)$.

That means $$1 = h_1(2(2) + 2(-1) - 1(1)) \implies h_1 = 1 $$ $$1 = h_2(1(2) + (-1) + 2(1)) \implies h_2 = \frac{1}{3} $$ $$1 = h_3(2(2) - 3(-1) - (1)) \implies h_3 = \frac{1}{6}$$

Hence the equations of the given point-coordinate transformation are

$$x’_1 = 2x_1 + 2x_2 - x_3$$ $$x’_2 = \frac{1}{3}(x_1 + x_2 + 2x_3)$$ $$x’_3 = \frac{1}{6}(2x_1 - 3x_2 - x_3)$$

As these are homogeneous coordinates, multiplying by a scalar will not change the coordinates. Hence, multiplying by 6 we get

$$x’_1 = 12x_1 + 12x_2 - 6x_3$$ $$x’_2 = 2x_1 + 2x_2 + 4x_3$$ $$x’_3 = 2x_1 - 3x_2 - x_3$$

(b) The line-coordinates of the axes in this case will be $$ x’_1: (0, 1, 1) \times (1, 1, 1) = [0, 1, -1] $$ $$ x’_2: (1, 1, 1) \times (1, 0, -1) = [-1, 2, -1] $$ $$ x’_3: (1, 0, -1) \times (0, 1, 1) = [1, -1, 1] $$

The equations expressing the new point-coordinates in terms of the old will be $$x’_1 = h_1(x_2 - x_3)$$ $$x’_2 = h_2(-x_1 + 2x_2 - x_3)$$ $$x’_3 = h_3(x_1 - x_2 + x_3)$$

provided $h_1, h_2, h_3$ are determined so that the new coordinates of the point $(1, 2, 0)$ are $(1, 1, 1)$.

That means $$1 = h_1(2 - 0) \implies h_1 = \frac{1}{2} $$ $$1 = h_2(-1 + 2(2) - 0) \implies h_2 = \frac{1}{3} $$ $$1 = h_3(1 - 2 + 0) \implies h_3 = -1$$

Hence the equations of the given point-coordinate transformation are

$$x’_1 = \frac{1}{2}(x_2 - x_3)$$ $$x’_2 = \frac{1}{3}(-x_1 + 2x_2 - x_3)$$ $$x’_3 = -1(x_1 - x_2 + x_3)$$

As these are homogeneous coordinates, multiplying by a scalar will not change the coordinates. Hence, multiplying by -6 we get

$$x’_1 = -3x_2 + 3x_3$$ $$x’_2 = 2x_1 - 4x_2 + 2x_3$$ $$x’_3 = 6x_1 - 6x_2 + 6x_3$$

8. Given $\lambda_1: x_2 + x_3 = 0$, $\lambda_2: x_1 + x_3 = 0$, $\lambda_3: x_1 + x_2 = 0$, and $U:(1, 0, 1)$. Find the equations of the point-coordinate transformation which assigns to $\lambda_1$, $\lambda_2$, $\lambda_3$ the new equations $x’_1 = 0$, $x’_2 = 0$, $x’_3 = 0$, respectively, and makes $U$ the new unit point. Find the equations of the line-coordinate transformation based on the triangle whose vertices are the points $\lambda_2\lambda_3$, $\lambda_1\lambda_3$, $\lambda_1\lambda_2$ and for which $x_1 + x_2 - 2x_3 = 0$ is the new unit line. What is the incidence condition for the new coordinate systems? What line must be chosen for the new unit line if the form of the incidence condition is to be preserved?

The equations expressing the new point-coordinates in terms of the old will be

$$x’_1 = h_1(x_2 + x_3)$$ $$x’_2 = h_2(x_1 + x_3)$$ $$x’_3 = h_3(x_1 + x_2)$$

provided $h_1, h_2, h_3$ are determined so that the new coordinates of the point $(1, 0, 1)$ are $(1, 1, 1)$.

That means $$1 = h_1(0 + 1) \implies h_1 = 1$$ $$1 = h_2(1 + 1) \implies h_2 = \frac{1}{2}$$ $$1 = h_3(1 + 0) \implies h_3 = 1$$

Hence the equations of the point-coordinate transformations are $$x’_1 = x_2 + x_3$$ $$x’_2 = \frac{1}{2}(x_1 + x_3)$$ $$x’_3 = x_1 + x_2$$

And the transformation matrices are

$$A = \begin{pmatrix} 0 & 1 & 1 \\ \frac{1}{2} & 0 & \frac{1}{2} \\ 1 & 1 & 0 \end{pmatrix}$$

$$A^{-1} = \begin{pmatrix} -0.5 & 1 & 0.5 \\ 0.5 & -1 & 0.5 \\ 0.5 & 1 & -0.5 \end{pmatrix}$$

The vertices of the new line-coordinate system will be $$\lambda_2\lambda_3: [1, 0, 1] \times [1, 1, 0] = (-1, 1, 1)$$ $$\lambda_1\lambda_3: [0, 1, 1] \times [1, 1, 0] = (-1, 1, -1)$$ $$\lambda_1\lambda_2: [0, 1, 1] \times [1, 0, 1] = (1, 1, -1)$$

The equations expressing the new line-coordinates in terms of the old will be

$$l’_1 = k_1(-l_1 + l_2 + l_3)$$ $$l’_2 = k_2(-l_1 + l_2 - l_3)$$ $$l’_3 = k_3(l_1 + l_2 - l_3)$$

provided $k_1, k_2, k_3$ are determined so that the new coordinates of the line $[1, 1, -2]$ are $[1, 1, 1]$.

That means $$1 = k_1(-1 + 1 + (-2)) \implies k_1 = -\frac{1}{2}$$ $$1 = k_2(-1 + 1 - (-2)) \implies k_2 = \frac{1}{2}$$ $$1 = k_3(1 + 1 - (-2)) \implies k_3 = \frac{1}{4}$$

Hence the equations of the line-coordinate transformations are $$l’_1 = -\frac{1}{2}(-l_1 + l_2 + l_3)$$ $$l’_2 = \frac{1}{2}(-l_1 + l_2 - l_3)$$ $$l’_3 = \frac{1}{4}(l_1 + l_2 - l_3)$$

And the transformation matrices are

$$B = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{4} & \frac{1}{4} & -\frac{1}{4} \end{pmatrix}$$

$$B^{-1} = \begin{pmatrix} 0 & -1 & 2 \\ -1 & 0 & 2 \\ -1 & -1 & 0 \end{pmatrix}$$

So, $(B^{-1})^TA^{-1}$ will be

$$ \begin{pmatrix} 0 & -1 & 2 \\ -1 & 0 & 2 \\ -1 & -1 & 0 \end{pmatrix}^T \begin{pmatrix} -0.5 & 1 & 0.5 \\ 0.5 & -1 & 0.5 \\ 0.5 & 1 & -0.5 \end{pmatrix} $$

$$ = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{pmatrix} $$

So the incidence condition will be

$$ l’_1x’_1 + 2l’_2x’_2 - 2l’_3x’_3 = 0$$

By Theorem 1, for the form of the incidence condition to be preserved, the unit line must be the polar of the new unit point with respect to the triangle with sides $\lambda_1$, $\lambda_2$, $\lambda_3$.

To find the polar, we must 1. Find the lines from the new unit point to each of the vertices of the triangle of reference. $$\lambda_2\lambda_3U: (-1, 1, 1) \times (1, 0, 1) = [1, 2, -1] $$ $$\lambda_1\lambda_3U: (-1, 1, -1) \times (1, 0, 1) = [1, 0, -1] $$ $$\lambda_1\lambda_2U: (1, 1, -1) \times (1, 0, 1) = [1, -2, -1] $$ 2. Find the harmonic conjugates of each of these lines with respect to the other two sides of the triangle of reference.
- Harmonic conjugate of [1, 2, -1] with respect to [1, 0, 1] and [1, 1, 0] is [3, 2, 1]
- Harmonic conjugate of [1, 0, -1] with respect to [0, 1, 1] and [1, 1, 0] is [1, 2, 1]
- Harmonic conjugate of [1, -2, -1] with respect to [0, 1, 1] and [1, 0, 1] is [1, 2, 3]
3. Find the intersections of each of these harmonic conjugates with the side opposite to the chosen vertex. $$ [3, 2, 1] \times [0, 1, 1] = (1, -3, 3) $$ $$ [1, 2, 1] \times [1, 0, 1] = (2, 0, -2) $$ $$ [1, 2, 3] \times [1, 1, 0] = (-3, 3, -1) $$ 4. Verify that they are indeed collinear $$ \begin{vmatrix} 1 & -3 & 3 \\ 2 & 0 & -2 \\ -3 & 3 & -1 \end{vmatrix} = 0 $$ 5. Find the line that determined by these three points. This should be the new unit line if we want to preserve the form of the incidence condition. $$ (1, -3, 3) \times (2, 0, -2) = [6, 8, 6] $$

Hence, the new unit line that preserves the form of the incidence condition is $3x_1 + 4x_2 + 3x_3 = 0$.

9. (a) It is desired to convert the equation of the parabola $y = x^2$ into the equation of a circle by suitably specializing a new coordinate system in which $x_1 = 0$, $x_1 + x_2 = 0$ and $ax_1 + ax_2 + ax_3 = 0$ are, respectively, the lines $x’_1 = 0$, $x’_2 = 0$, and $x’_3 = 0$. Determine what relations $a_1$, $a_2$, $a_3$ must satisfy and what the new unit point must be if this is to be accomplished.
(b) Work part (a) if the equation of the hyperbola $xy = 1$ is to be converted into the equation of a circle.

The equations expressing the new point-coordinates in terms of the old will be

$$x’_1 = h_1(x_1)$$ $$x’_2 = h_2(x_1 + x_2)$$ $$x’_3 = h_3(a_1x_1 + a_2x_2 + a_3x_3)$$

So the matrix of transformation will be

$$ A = \begin{pmatrix} h_1 & 0 & 0 \\ h_2 & h_2 & 0 \\ h_3a_1 & h_3a_2 & h_3a_3 \end{pmatrix} $$

such that

$$ X’ = AX $$

The inverse transformation will then be

$$ A^{-1} = \begin{pmatrix} \frac{1}{h_1} & 0 & 0 \\ -\frac{1}{h_1} & \frac{1}{h_2} & 0 \\ \frac{(a_2 - a_1)}{h_1a_3} & -\frac{a_2}{h_2a_3} & \frac{1}{h_3a_3} \end{pmatrix} $$

(a) First, let’s write the equation of the parabola in homogeneous coordinates.

$$ x_1^2 = x_2x_3 $$

Applying the point-coordinate transformation to the equation of the parabola, we get

$$ (\frac{x’_1}{h_1})^2 = (\frac{x’_2}{h_2} - \frac{x’_1}{h_1})(\frac{(a_2 - a_1)x’_1}{h_1a_3} - \frac{x’_2a_2}{h_2a_3} + \frac{x’_3}{h_3a_3}) $$

$$\implies (\frac{x’_1}{h_1})^2 = (\frac{x’_2h_1 - x’_1h_2}{h_1h_2})(\frac{(a_2 - a_1)x’_1h_2h_3 - x’_2a_2h_1h_3 + x’_3h_1h_2}{h_1h_2h_3a_3}) $$

$$\implies (x’_1h_2)^2h_3a_3 = (x’_2h_1 - x’_1h_2)((a_2 - a_1)x’_1h_2h_3 - x’_2a_2h_1h_3 + x’_3h_1h_2) $$

$$\implies (a_3 + a_2 - a_1)h_3(h_2x’_1)^2 + a_2h_3(h_1x’_2)^2 + \\ (a_1 - 2a_2)h_1h_2h_3x’_1x’_2 + h_1h_2^2x’_1x’_3 - h_1^2h_2x’_2x’_3 = 0 $$

Choosing $x’_3$ as the ideal line and setting $x = \frac{x’_1}{x’_3}$, $y = \frac{x’_2}{x’_3}$ we get

$$(a_3 + a_2 - a_1)h_2^2h_3x^2 + a_2h_1^2h_3y^2 + (a_1 - 2a_2)h_1h_2h_3xy + h_1h_2^2x - h_1^2h_2y = 0 $$

For this to be a circle, the $xy$ term must be zero and the coefficients of the $x^2$ and $y^2$ terms must be equal.

As none of the $h_i$s can be zero, that means,

$$a_1 = 2a_2$$ and, $$(a_3 + a_2 - a_1)h_2^2 = a_2h_1^2$$ $$ \implies (a_3 + a_2 - 2a_2)h_2^2 = a_2h_1^2$$ $$ \implies \frac{(a_3 - a_2)}{a_2} = \frac{h_1^2}{h_2^2}$$

These are the two constraints we will use to derive the relationship between the $a_i$s and the unit point.

From the equations of transformation, we know that $$x’_1 = h_1(x_1)$$ $$x’_2 = h_2(x_1 + x_2)$$

Squaring these two and dividing one by the other we get

$$\frac{(x’_1)^2}{(x’_2)^2} = \frac{h_1^2x_1^2}{h_2^2(x_1 + x_2)^2}$$

Using the constraint between $h_1$ and $h_2$ we derived earlier, we get

$$\frac{(x’_1)^2}{(x’_2)^2} = \frac{(a_3 - a_2)}{a_2}\frac{x_1^2}{(x_1 + x_2)^2}$$

Substituting coordinates of the unit point in the new coordinates we will get the equation for the unit point in the old coordinates $$\frac{(a_3 - a_2)}{a_2}\frac{x_1^2}{(x_1 + x_2)^2} = 1$$ $$(a_3 - a_2)x_1^2 = a_2(x_1 + x_2)^2$$ $$ \implies (a_3 - 2a_2)x_1^2 - a_2(x_2^2 + 2x_1x_2) = 0$$

So, the third axis (also the ideal line of the specialization) of the new triangle of reference can be any line in the envelope $a_1 = 2a_2$ provided the unit point is a point on the conic $(a_3 - 2a_2)x_1^2 - a_2(x_2^2 + 2x_1x_2) = 0$.

(b) First, let’s write the equation of the hyperbola in homogeneous coordinates.

$$ x_1x_2 = x_3^2 $$

Applying the point-coordinate transformation to the equation of the hyperbola, we get

$$(\frac{x’_1}{h_1})(\frac{x’_2}{h_2} - \frac{x’_1}{h_1}) = (\frac{(a_2 - a_1)x’_1}{h_1a_3} - \frac{x’_2a_2}{h_2a_3} + \frac{x’_3}{h_3a_3})^2$$

$$ \implies \frac{(x’_2h_1 - x’_1h_2)x’_1}{h_1^2h_2} = (\frac{(a_2 - a_1)x’_1h_2h_3 - x’_2a_2h_1h_3 + x’_3h_1h_2}{h_1h_2h_3a_3})^2 $$

$$ \implies (x’_2h_1 - x’_1h_2)x’_1 = \frac{((a_2 - a_1)x’_1h_2h_3 - x’_2a_2h_1h_3 + x’_3h_1h_2)^2}{h_2h_3^2a_3^2} $$

$$ \implies (x’_2h_1 - x’_1h_2)x’_1h_2h_3^2a_3^2 = ((a_2 - a_1)x’_1h_2h_3 - x’_2a_2h_1h_3 + x’_3h_1h_2)^2 $$

$$ \implies (h_1h_2x’_3)^2 + (a_2h_1h_3x’_2)^2 + (a_3^2 + a_2^2 + a_1^2 - 2a_1a_2)(h_2h_3x’_1)^2 - \\ (a_3^2 + 2a_2^2 - 2a_1a_2)h_1h_2h_3^2x’_1x’_2 - 2a_2h_1^2h_2h_3x’_2x’_3 - \\ (2a_1 - 2a_2)h_1h_2^2h_3x’_1x’_3 = 0$$

Choosing $x’_3$ as the ideal line and and setting $x = \frac{x’_1}{x’_3}$, $y = \frac{x’_2}{x’_3}$ we get

$$ \implies (h_1h_2)^2 + (a_2h_1h_3y)^2 + (a_3^2 + a_2^2 + a_1^2 - 2a_1a_2)(h_2h_3x)^2 - \\ (a_3^2 + 2a_2^2 - 2a_1a_2)h_1h_2h_3^2xy - 2a_2h_1^2h_2h_3y - \\ (2a_1 - 2a_2)h_1h_2^2h_3x = 0$$

For this to be a circle, the $xy$ term must be zero and the coefficients of the $x^2$ and $y^2$ terms must be equal.

As none of the $h_i$s can be zero, that means, $$a_3^2 + 2a_2^2 - 2a_1a_2 = 0$$ and, $$(a_2h_1)^2 = (a_3^2 + a_2^2 + a_1^2 - 2a_1a_2)(h_2)^2$$ $$\implies (a_2h_1)^2 = (-2a_2^2 + 2a_1a_2 + a_2^2 + a_1^2 - 2a_1a_2)(h_2)^2$$ $$\implies (a_2h_1)^2 = (a_1^2 - a_2^2)(h_2)^2$$ $$\implies (\frac{h_1}{h_2})^2 = \frac{a_1^2 - a_2^2}{a_2^2}$$

These are the two constraints we will use to derive the relationship between the $a_i$s and the unit point.

From the equations of transformation, we know that $$\frac{(x’_1)^2}{(x’_2)^2} = \frac{h_1^2x_1^2}{h_2^2(x_1 + x_2)^2}$$

Using the constraint between $h_1$ and $h_2$ we derived earlier, we get $$\frac{(x’_1)^2}{(x’_2)^2} = \frac{(a_1^2 - a_2^2)x_1^2}{a_2^2(x_1 + x_2)^2}$$

Substituting coordinates of the unit point in the new coordinates we will get the equation for the unit point in the old coordinates $$\frac{(a_1^2 - a_2^2)x_1^2}{a_2^2(x_1 + x_2)^2} = 1$$ $$\implies (a_1^2 - a_2^2)x_1^2 - a_2^2(x_1 + x_2)^2 = 0$$ $$\implies a_1^2x_1^2 = a_2^2(2x_1^2 + x_2^2 + 2x_1x_2)$$

So, the third axis (also the ideal line of the specialization) of the new triangle of reference can be any line in the envelope $a_3^2 + 2a_2^2 - 2a_1a_2 = 0$ provided the unit point is a point on the conic $a_1^2x_1^2 = a_2^2(2x_1^2 + x_2^2 + 2x_1x_2) = 0$.

10. Discuss the possibility of introducing new homogeneous coordinates in $E_2$ in the following way. Let $\bigtriangleup A_1A_2A_3$ be an arbitrary triangle, let $P$ be an arbitrary point, and associate with $P$ the ordered triple $(kx’_1, kx’_2, kx’_3)$, where $x’_i$ is the area of $\bigtriangleup PA_jA_k$. Show that this procedure is a special case of the general process of introducing new homogeneous coordinates which we developed in this section. Can this process be extended to $E_2^+$

A major part of the solution is deciding which formula to use to find the area of a triangle. There are several ways¹ to do so but as we’re dealing with coordinates, the most fitting way would be to use the determinant-based shoelace formula²³.

Let’s represent the cartesian coordinates of the vertices of the triangle as $A_i: (a_{ix}, a_{iy})$ and that of the general point $P$ as $(p_x, p_y)$.

According to the shoelace formula for triangles, the area of the triangles $\bigtriangleup PA_jA_k$ would be

$$Area(\bigtriangleup PA_jA_k) = \frac{1}{2}\begin{vmatrix} a_{jx} & a_{kx} & p_x \\ a_{jy} & a_{ky} & p_y \\ 1 & 1 & 1 \end{vmatrix}$$ $$= \frac{1}{2}|(a_{jy} - a_{ky})p_x + (a_{kx} - a_{jx})p_y + (a_{jx}a_{ky} - a_{jy}a_{kx})|$$

This can also be written as

$$\frac{1}{2} \begin{pmatrix} a_{jy} - a_{ky} & a_{kx} - a_{jx} & a_{jx}a_{ky} - a_{jy}a_{kx} \end{pmatrix} \begin{pmatrix} p_x \\ p_y \\ 1 \end{pmatrix}$$

Now, the line determined by $A_jA_k$ is $A_j \times A_k = [a_{jy} - a_{ky}, a_{kx} - a_{jx}, a_{jx}a_{ky} - a_{jy}a_{kx}]$.

So the area $\bigtriangleup PA_jA_k$ can be re-written as $$\frac{1}{2} \begin{pmatrix} A_j \times A_k \end{pmatrix} \begin{pmatrix} p_x \\ p_y \\ 1 \end{pmatrix}$$

Hence, the new point-coordinates of a general point P in homogeneous coordinates can be expressed as $$P’ = AP$$ where $$A = \frac{1}{2}\begin{pmatrix} A_2 \times A_3 \\ A_3 \times A_1 \\ A_1 \times A_2 \end{pmatrix}$$ and $$P = \begin{pmatrix} p_x \\ p_y \\ 1 \end{pmatrix}$$

This is exactly the same as following the general process of introducing new homogeneous coordinates developed in this section with $\bigtriangleup A_1A_2A_3$ being the new triangle of reference and the centroid of the triangle being the new unit point as the centroid divides the triangle into three triangles having equal area.

We can transform ideal points, and hence $E_2^+$, using this transformation, by setting the third coordinate in $P$ to be zero.

In fact, this representation of coordinates is a special case of homogeneous coordinates called homogeneous barycentric coordinates⁴. Barycentric coordinates⁵ were the very first kind of homogeneous coordinates to be used in projective geometry. Barycentre is a word with Greek origins that comes from astronomy and means the center of mass of multiple bodies that orbit one another. In geometry it is a synonym for the centroid of a geometric shape. It was introduced by August Ferdinand Möbius, a German mathematician and astronomer, who also introduced homogeneous coordinates into projective geometry⁶. I think this exercise pays homage to him and also introduces the reader to the concept of barycentric coordinates which are heavily used in computer graphics.

References

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