Question 2 of 7: Two-dimensional geometric transformations

Question 2: Two-dimensional geometric transformations (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Part (a): a straight line segment with end points $P_1 = (0,\;0)$ and $P_2 = (2,\;3)$. Part (b): a segment with end points $Q_1 = (1,\;1)$ and $Q_2 = (1,\;3)$, to be carried to $Q_1' = (0,\;1)$ and $Q_2' = (0,\;5)$.

Find. The matrix representation of each line, the image of the part-(a) line under each of the four transformations, and a sequence of elementary transformations (with their matrices) that produces the part-(b) movement.

Approach. Write both end points as homogeneous column vectors so that scaling, rotation and translation are all $3\times 3$ matrix multiplications, apply each elementary matrix to the point matrix, and for part (b) build the required mapping as translate → scale → translate-back, then collapse the three into one concatenated matrix.

1. Part (a) — write the line in homogeneous matrix notation. A translation is not a linear map on $(x,\,y)$, so plane geometry in CAD is carried out in homogeneous coordinates: the point $(x,\,y)$ becomes the column $[x\;\;y\;\;1]^{T}$ and every transformation becomes a single $3\times 3$ matrix. Storing the two end points as the columns of one matrix lets both be transformed at once: $$\mathbf{P}=\begin{bmatrix} 0 & 2 \\ 0 & 3 \\ 1 & 1 \end{bmatrix} \qquad\text{(column 1 is } P_1\text{, column 2 is } P_2\text{)}$$ The image of the line under a transformation matrix $\mathbf{T}$ is then simply $\mathbf{P}' = \mathbf{T}\,\mathbf{P}$, and the transformed end points are read off as the columns of $\mathbf{P}'$.
2. (i) Scale uniformly by 2.0. A scaling about the origin multiplies each coordinate by its scale factor, so with $s_x = s_y = 2.0$ $$\mathbf{S}(2,2)=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \qquad \mathbf{S}\,\mathbf{P}=\begin{bmatrix} 0 & 4 \\ 0 & 6 \\ 1 & 1 \end{bmatrix}$$ The end points move to $\boxed{(0,\,0)\ \text{and}\ (4,\,6)}$. The origin is the fixed point of the transformation, so $P_1$ does not move; the line keeps its direction and doubles in length, from $\sqrt{13}=3.606$ to $\sqrt{52}=7.211$ units.
3. (ii) Scale 3.0 in x and 2.0 in y. Differential scaling uses the same matrix with unequal diagonal entries: $$\mathbf{S}(3,2)=\begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \qquad \mathbf{S}\,\mathbf{P}=\begin{bmatrix} 0 & 6 \\ 0 & 6 \\ 1 & 1 \end{bmatrix}$$ giving $\boxed{(0,\,0)\ \text{and}\ (6,\,6)}$. Unlike (i) this is not a similarity transformation: the slope changes from $3/2$ to $1$, so the line now lies at 45°. Differential scaling distorts angles, which is why a CAD system applies it to a whole model only with care.
4. (iii) Rotate 45° about the origin. A counter-clockwise rotation through $\theta$ has the matrix $$\mathbf{R}(\theta)=\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}, \qquad \cos 45^{\circ}=\sin 45^{\circ}=0.7071$$ Applying it to the second column, $x' = 2(0.7071) - 3(0.7071) = -0.7071$ and $y' = 2(0.7071) + 3(0.7071) = 3.5355$, so the image is $$\boxed{(0,\,0)\ \text{and}\ (-0.707,\;3.536)}$$ As a check, rotation is an orthogonal transformation and must preserve length: $\sqrt{0.7071^{2}+3.5355^{2}} = 3.606 = \sqrt{13}$, which matches the original length exactly. The end point has swung from the first quadrant into the second, as it must, since the original line lies at $\arctan(3/2) = 56.31^{\circ}$ and $56.31 + 45 = 101.31^{\circ}$.
5. (iv) Translate 2.0 units in x and 2.0 units in y. Translation is the reason for the homogeneous third row: the shift appears in the third column, $$\mathbf{T}(2,2)=\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}, \qquad \mathbf{T}\,\mathbf{P}=\begin{bmatrix} 2 & 4 \\ 2 & 5 \\ 1 & 1 \end{bmatrix}$$ so the end points become $\boxed{(2,\,2)\ \text{and}\ (4,\,5)}$. Both length and direction are unchanged — the whole segment simply shifts along the vector $(2,\,2)$.

6. Part (b)(i) — diagnose what the required movement actually does. The starting segment runs from $(1,1)$ to $(1,3)$: it is vertical, two units long, at $x = 1$. The target runs from $(0,1)$ to $(0,5)$: still vertical, but four units long and at $x = 0$. So the movement is a doubling of length in the y-direction with the lower end point held fixed at $y = 1$, combined with a one-unit shift to the left. A scaling matrix always works about the origin, so the y-scaling cannot be applied directly — it would move the point $(1,1)$ to $(1,2)$. The standard device is to translate the fixed point onto the origin, scale there, then translate back.
7. Part (b)(i) — state the sequence. Three elementary transformations, in this order:
     Step 1. Translate the line by $-1$ in y, carrying the lower end point $(1,1)$ onto the x-axis at $(1,0)$.
     Step 2. Scale by a factor of $2$ in the y-direction (and $1$ in x) about the origin, which stretches the segment from two units to four while the lower end stays on the axis.
     Step 3. Translate by $-1$ in x and $+1$ in y, which restores the lower end point to $y = 1$ and simultaneously moves the line onto the axis $x = 0$.
   The x-shift is folded into the final translation rather than given a step of its own, because the x-scale factor is unity and translations commute in x.
8. Part (b)(ii) — write the three matrices. In the order applied, $$\mathbf{T}_1=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}, \qquad \mathbf{S}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \qquad \mathbf{T}_2=\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$$ where $\mathbf{T}_1$ translates by $(0,-1)$, $\mathbf{S}$ scales by $(1,2)$ about the origin, and $\mathbf{T}_2$ translates by $(-1,+1)$.
9. Concatenate and verify. Because the transformations are applied to the point on the right, they concatenate right-to-left: $$\mathbf{M}=\mathbf{T}_2\,\mathbf{S}\,\mathbf{T}_1 =\begin{bmatrix} 1 & 0 & -1 \\ 0 & 2 & -1 \\ 0 & 0 & 1 \end{bmatrix}$$ Applying $\mathbf{M}$ to the two end points, $$\mathbf{M}\begin{bmatrix} 1 & 1 \\ 1 & 3 \\ 1 & 1 \end{bmatrix} =\begin{bmatrix} 0 & 0 \\ 1 & 5 \\ 1 & 1 \end{bmatrix} \;\Longrightarrow\; \boxed{(0,\,1)\ \text{and}\ (0,\,5)}$$ which is the required result. Concatenating first and transforming once is what a real graphics pipeline does: one matrix multiply per vertex instead of three.