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 $c = 4$

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.