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The problem asks whether the given pairs of lines are perpendicular to each other or not. For lines to be perpendicular, the product of their slopes must be \(-1\).

Here are the steps to check if the lines in each case are perpendicular:

1. **Find the slopes of the lines**:  
   The slope of a line between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
   \[
   m = \frac{y_2 - y_1}{x_2 - x_1}
   \]
   
2. **Check if the slopes are perpendicular**:  
   If the product of the slopes of two lines is \(-1\), then the lines are perpendicular.

Now, let's calculate the slopes for each option.

---

### a. Points: 
- Line 1: \( A(3, 1), B(4, 3) \)
- Line 2: \( C(1, -3), D(0, -2) \)

#### Line 1 slope:
\[
m_1 = \frac{3 - 1}{4 - 3} = \frac{2}{1} = 2
\]
#### Line 2 slope:
\[
m_2 = \frac{-2 + 3}{0 - 1} = \frac{1}{-1} = -1
\]
\[
m_1 \times m_2 = 2 \times (-1) = -2 \quad (\text{Not perpendicular})
\]

### b. Points: 
- Line 1: \( A(-1, -2), B(4, 5) \)
- Line 2: \( C(2, -5), D(0, 0) \)

#### Line 1 slope:
\[
m_1 = \frac{5 + 2}{4 + 1} = \frac{7}{5}
\]
#### Line 2 slope:
\[
m_2 = \frac{0 + 5}{0 - 2} = \frac{5}{-2} = -\frac{5}{2}
\]
\[
m_1 \times m_2 = \frac{7}{5} \times -\frac{5}{2} = -\frac{35}{10} = -3.5 \quad (\text{Not perpendicular})
\]

### c. Points:
- Line 1: \( A(4, -5), B(0, -2) \)
- Line 2: \( C(0, 0), D(3, 4) \)

#### Line 1 slope:
\[
m_1 = \frac{-2 + 5}{0 - 4} = \frac{3}{-4} = -\frac{3}{4}
\]
#### Line 2 slope:
\[
m_2 = \frac{4 - 0}{3 - 0} = \frac{4}{3}
\]
\[
m_1 \times m_2 = -\frac{3}{4} \times \frac{4}{3} = -1 \quad (\text{Perpendicular!})
\]

### d. Points: 
- Line 1: \( A(0, 2), B(4, 4) \)
- Line 2: \( C(4, -3), D(0, 0) \)

#### Line 1 slope:
\[
m_1 = \frac{4 - 2}{4 - 0} = \frac{2}{4} = \frac{1}{2}
\]
#### Line 2 slope:
\[
m_2 = \frac{0 + 3}{0 - 4} = \frac{3}{-4} = -\frac{3}{4}
\]
\[
m_1 \times m_2 = \frac{1}{2} \times -\frac{3}{4} = -\frac{3}{8} \quad (\text{Not perpendicular})
\]

### e. Points: 
- Line 1: \( A(-2, 0), B(10, 8) \)
- Line 2: \( C(2, 3), D(6, -3) \)

#### Line 1 slope:
\[
m_1 = \frac{8 - 0}{10 + 2} = \frac{8}{12} = \frac{2}{3}
\]
#### Line 2 slope:
\[
m_2 = \frac{-3 - 3}{6 - 2} = \frac{-6}{4} = -\frac{3}{2}
\]
\[
m_1 \times m_2 = \frac{2}{3} \times -\frac{3}{2} = -1 \quad (\text{Perpendicular!})
\]

---

### Conclusion:
- Option **c** and **e** are perpendicular.

Would you like further details or have any questions?

---

Here are five related questions to deepen your understanding:
1. How do you find the equation of a line given two points?
2. What is the geometric meaning of the product of slopes being \(-1\

Tentukanlah apakah garis-garis berikut ini tegak lurus satu sama lain atau tidak: a. A(3, 1), B(4, 3) dan C(1, -3), D(0, -2); b. A(-1, -2), B(4, 5) dan C(2, -5), D(0, 0); c. A(4, -5), B(0, -2) dan C(0, 0), D(3, 4); d. A(0, 2), B(4, 4) dan C(4, -3), D(0, 0); e. A(-2, 0), B(10, 8) dan C(2, 3), D(6, -3)

Solve the problem of determining if pairs of lines are perpendicular by calculating the slopes of the lines and checking if the product of the slopes is -1. This problem is suitable for high school students learning about analytic geometry.

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