Solved Numericals

Q1. DDA Line from (3,2) to (11,7)

Given: $P_1(3,2), P_2(11,7)$

Calculate Slope:
$dx = 11 - 3 = 8$
$dy = 7 - 2 = 5$
Determine Steps:
Since $|dx| > |dy|$ ($8 > 5$), $Steps = 8$.
Calculate Increment:
$X_{inc} = \frac{8}{8} = 1$
$Y_{inc} = \frac{5}{8} = 0.625$

k	X	Y	Plot (Round)
0	3	2	(3, 2)
1	4	2.625	(4, 3)
2	5	3.25	(5, 3)
3	6	3.875	(6, 4)
4	7	4.5	(7, 5)
5	8	5.125	(8, 5)
6	9	5.75	(9, 6)
7	10	6.375	(10, 6)
8	11	7.0	(11, 7)

Q2. DDA Line from (1,1) to (4,9)

Given: $P_1(1,1), P_2(4,9)$

$dx = 4 - 1 = 3$
$dy = 9 - 1 = 8$
Since $|dy| > |dx|$, $Steps = 8$.
$X_{inc} = \frac{3}{8} = 0.375$, $Y_{inc} = \frac{8}{8} = 1$.

k	X	Y	Plot (Round)
0	1	1	(1, 1)
1	1.375	2	(1, 2)
2	1.75	3	(2, 3)
3	2.125	4	(2, 4)
4	2.5	5	(3, 5)
5	2.875	6	(3, 6)
6	3.25	7	(3, 7)
7	3.625	8	(4, 8)
8	4.0	9	(4, 9)

Q3. Bresenham Line from (2,3) to (12,8)

Given: $P_1(2,3), P_2(12,8)$

$dx = 10, dy = 5$.
Constants: $2dy = 10$, $2dy - 2dx = 10 - 20 = -10$.
Initial Decision: $P_0 = 2dy - dx = 10 - 10 = 0$.

k	$P_k$	Plot $(x_{k+1}, y_{k+1})$	Update Rule
0	0	(2, 3)	Start
1	0	(3, 4)	Since $P \ge 0$, Inc Both, $P = P - 10$ (-10)
2	-10	(4, 4)	Since $P < 0$, Inc X, $P=P + 10$ (0)
3	0	(5, 5)	Since $P \ge 0$, Inc Both, $P = -10$
4	-10	(6, 5)	$P < 0$, Inc X, $P=0$
5	0	(7, 6)	$P \ge 0$, Inc Both, $P=-10$
6	-10	(8, 6)	$P < 0$, Inc X, $P=0$
7	0	(9, 7)	$P \ge 0$, Inc Both, $P=-10$
8	-10	(10, 7)	$P < 0$, Inc X, $P=0$
9	0	(11, 8)	$P \ge 0$, Inc Both, end

Q4. Mid-Point Circle (r = 6) - First 8 Points

Constraint: Plot symmetric points.

Start $(0, 6)$.
$P_0 = 1 - r = 1 - 6 = -5$.

Quadrant 1 Octant 2 Calculation (x starts at 0, y at r):

k	$P_k$	Next Point	$P_{k+1}$ Formula
0	-5	(1, 6)	$P + 2x + 1 = -5 + 2(1) + 1 = -2$
1	-2	(2, 6)	$P + 2x + 1 = -2 + 2(2) + 1 = 3$
2	3	(3, 5)	$P + 2x - 2y + 1 = 3 + 6 - 10 + 1 = 0$
3	0	(4, 4)	End ($x=y$)

The First 8 Symmetric Points (from (1,6)):

            mindmap
              root((Symmetry))
                Point (1,6)
                Review
                  (1, 6)
                  (6, 1)
                  (1, -6)
                  (6, -1)
                  (-1, 6)
                  (-6, 1)
                  (-1, -6)
                  (-6, -1)

Q5. Circle (r = 10) - First 5 Points

Start: $(0, 10)$, $P_0 = 1 - 10 = -9$.

Pt 1: (0, 10). $P < 0 \implies$ Next (1, 10). $P=-9 + 3=-6$.
Pt 2: (1, 10). $P < 0 \implies$ Next (2, 10). $P=-6 + 5=-1$.
Pt 3: (2, 10). $P < 0 \implies$ Next (3, 10). $P=-1 + 7=6$.
Pt 4: (3, 10). $P \ge 0 \implies$ Next (4, 9). $P = 6 + 2(4) - 2(9) + 1 = -3$.
Pt 5: (4, 9).

Q6. Translate Triangle A(1,2), B(4,3), C(6,1)

Vector: $T_x = 2, T_y = 5$. Formula: $x' = x + T_x, y' = y + T_y$.

$A(1,2) \to (1+2, 2+5) \to \mathbf{A'(3, 7)}$
$B(4,3) \to (4+2, 3+5) \to \mathbf{B'(6, 8)}$
$C(6,1) \to (6+2, 1+5) \to \mathbf{C'(8, 6)}$

Q7. Scale Rectangle by $S_x=2, S_y=3$

Given: $P(2,2), Q(4,2), R(4,5), S(2,5)$. Formula: $x' = x \cdot S_x, y' = y \cdot S_y$.

$P(2,2) \to (2 \cdot 2, 2 \cdot 3) \to \mathbf{P'(4, 6)}$
$Q(4,2) \to (4 \cdot 2, 2 \cdot 3) \to \mathbf{Q'(8, 6)}$
$R(4,5) \to (4 \cdot 2, 5 \cdot 3) \to \mathbf{R'(8, 15)}$
$S(2,5) \to (2 \cdot 2, 5 \cdot 3) \to \mathbf{S'(4, 15)}$

Q8. Rotate Point (5,2) by 90°

About Origin:

$$x' = x \cos 90^\circ - y \sin 90^\circ = x(0) - y(1) = -y$$ $$y' = x \sin 90^\circ + y \cos 90^\circ = x(1) + y(0) = x$$

Substitution:

$x = 5, y = 2$
$x' = -2$
$y' = 5$
Result: $\mathbf{(-2, 5)}$

Q9. Rotate Triangle A(1,1), B(2,3), C(3,1) by 45°

$\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707$

Formulas:

$x' = 0.707x - 0.707y$
$y' = 0.707x + 0.707y$

Calculation:

Point	Calculation	Result (Approx)
A(1,1)	$x' = 0.707(1-1) = 0$ $y' = 0.707(1+1) = 1.414$	$\mathbf{A'(0, 1.41)}$
B(2,3)	$x' = 0.707(2-3) = -0.707$ $y' = 0.707(2+3) = 3.535$	$\mathbf{B'(-0.70, 3.53)}$
C(3,1)	$x' = 0.707(3-1) = 1.414$ $y' = 0.707(3+1) = 2.828$	$\mathbf{C'(1.41, 2.82)}$

Q10. Reflect Point (4,5) in Y-axis

Concept: Mirroring across the vertical axis flips the X sign.

            graph LR
                A((4, 5)) -- "Reflect Y" --> B((\-4, 5))

Formula: $x' = -x, \quad y' = y$

Result: $\mathbf{(-4, 5)}$