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Let's work through the mathematical problems shown in the image step by step:

### Question 5:
**Dapatkan koordinat semua titik \( P \) pada sumbu-\( x \) sehingga garis yang melalui \( A(1, 2) \) dan \( P \) tegak lurus ke garis yang melalui \( B(8, 3) \) dan \( P \).**

This question asks to find the coordinates of points \( P \) on the x-axis such that the line through \( A(1, 2) \) and \( P \) is perpendicular to the line through \( B(8, 3) \) and \( P \).

1. **Coordinates of \( P \) on the x-axis**: Since \( P \) lies on the x-axis, its coordinates are \( P(p, 0) \).

2. **Slope of line \( AP \)**: The slope of the line passing through points \( A(1, 2) \) and \( P(p, 0) \) is:
   \[
   m_{AP} = \frac{0 - 2}{p - 1} = \frac{-2}{p - 1}
   \]

3. **Slope of line \( BP \)**: The slope of the line passing through points \( B(8, 3) \) and \( P(p, 0) \) is:
   \[
   m_{BP} = \frac{0 - 3}{p - 8} = \frac{-3}{p - 8}
   \]

4. **Perpendicular lines**: For two lines to be perpendicular, the product of their slopes must be \( -1 \):
   \[
   m_{AP} \cdot m_{BP} = -1
   \]
   Substituting the slopes:
   \[
   \frac{-2}{p - 1} \cdot \frac{-3}{p - 8} = -1
   \]
   Simplifying:
   \[
   \frac{6}{(p - 1)(p - 8)} = -1
   \]
   Multiply both sides by \( (p - 1)(p - 8) \):
   \[
   6 = -(p - 1)(p - 8)
   \]
   Expanding the right-hand side:
   \[
   6 = -(p^2 - 9p + 8)
   \]
   Rearranging the equation:
   \[
   p^2 - 9p + 14 = 0
   \]
   Solve using the quadratic formula:
   \[
   p = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(14)}}{2(1)} = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2}
   \]
   \[
   p = \frac{9 \pm 5}{2}
   \]
   Therefore:
   \[
   p = \frac{9 + 5}{2} = 7 \quad \text{or} \quad p = \frac{9 - 5}{2} = 2
   \]

   So, the possible coordinates of \( P \) are \( (7, 0) \) and \( (2, 0) \).

---

### Question 6:
**Dapatkan kemiringan garis dan perpotongannya dengan sumbu-\( y \) dari persamaan linear berikut:**

#### (a) \( 3x - 4y = 12 \)
1. Rearrange into slope-intercept form \( y = mx + c \):
   \[
   -4y = -3x + 12 \quad \Rightarrow \quad y = \frac{3}{4}x - 3
   \]
   **Slope** \( m = \frac{3}{4} \), **y-intercept** \( c = -3 \).

#### (b) \( \frac{x}{3} + \frac{y}{4} = 1 \)
1. Rearrange into slope-intercept form \( y = mx + c \):
   \[
   \frac{y}{4} = -\frac{x}{3} + 1 \quad \Rightarrow \quad y = -\frac{4}{3}x + 4
   \]
   **Slope** \( m = -\frac{4}{3} \), **y-intercept** \( c = 4 \).

#### (c) \( px + qy + r = 0 \) where \( q \neq 0 \)
1. Rearrange into slope-intercept form \( y = mx + c \):
   \[
   qy = -px - r \quad \Rightarrow \quad y = -\frac{p}{q}x - \frac{r}{q}
   \]
   **Slope** \( m = -\frac{p}{q} \), **y-intercept** \( c = -\frac{r}{q} \).

#### (d) \( y = 4 - \frac{1}{3}x \)
This is already in slope-intercept form \( y = mx + c \).
- **Slope** \( m = -\frac{1}{3} \)
- **y-intercept** \( c = 4 \)

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Would you like further explanation on any of these problems or the next questions? Here are 5 related questions to expand on:

1. How do we determine the equation of a line perpendicular to a given line?
2. Can you explain how to use the quadratic formula step by step?
3. What is the relationship between slopes of parallel lines?
4. How can we convert a general form equation to slope-intercept form?
5. What is the geometric interpretation of a slope and intercept?

**Tip**: When solving for the intersection of two lines, you can set the equations equal to each other and solve for the variables to find the point of intersection.

Dapatkan koordinat semua titik P pada sumbu-x sehingga garis yang melalui A(1, 2) dan P tegak lurus ke garis yang melalui B(8, 3) dan P. Dapatkan kemiringan garis dan perpotongannya dengan sumbu-y dari persamaan linear berikut ini.

Solve to find the coordinates of point P on the x-axis so that the line through A(1, 2) and P is perpendicular to the line through B(8, 3) and P. This problem explores slope relationships and requires solving a quadratic equation. Suitable for Grades 9-11.

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