https://file.aiquickdraw.com/mathbot/file/1730117732816fvugd81y.jpg

Let's analyze the question from the image:

The problem involves a Cartesian plane where several lines, including \( PQ \), \( QR \), and \( RS \), are given. We're asked to:

1. **(a)** Find the equation of the line \( QR \).
2. **(b)** Determine the equation of the line \( RS \).

### Information from the question:
- The **equation of line \( PQ \)** is:  
  \[ 3x + 2y = 6 \]
- Point \( S \) has coordinates:  
  \( S(-3, 10) \)
- Point \( Q \) lies on the **y-axis** (so its \( x \)-coordinate is 0).
- The line \( PQ \) is **parallel to the line \( RS \)**, meaning both lines will have the **same slope**.
- The line \( QR \) is **parallel to the x-axis** (so it is a horizontal line).

---

### Solution:

#### (a) Finding the equation of the line \( QR \):
Since the line \( QR \) is **parallel to the x-axis**, the equation of the line will have the form:  
\[
y = c
\]
where \( c \) is the y-coordinate of point \( Q \).  
Since point \( Q \) lies on the **y-axis** (at \( x = 0 \)) and also on the line \( PQ \), we can substitute \( x = 0 \) into the equation of \( PQ \) to determine the y-coordinate of \( Q \).

\[
3(0) + 2y = 6 \implies 2y = 6 \implies y = 3
\]
Thus, the coordinates of \( Q \) are \( (0, 3) \), and the equation of the horizontal line \( QR \) is:
\[
y = 3
\]

---

#### (b) Determining the equation of the line \( RS \):
Since **line \( RS \) is parallel to \( PQ \)**, it will have the **same slope** as \( PQ \). Let's find the slope of the line \( PQ \) from its equation.

The equation of \( PQ \) is:
\[
3x + 2y = 6
\]
Rearrange it into slope-intercept form \( y = mx + c \):
\[
2y = -3x + 6 \implies y = -\frac{3}{2}x + 3
\]
Thus, the **slope of \( PQ \)** (and also of \( RS \)) is:
\[
m = -\frac{3}{2}
\]

Now, we know that the line \( RS \) passes through the point \( S(-3, 10) \). Using the **point-slope form** of a line’s equation:
\[
y - y_1 = m(x - x_1)
\]
where \( (x_1, y_1) = (-3, 10) \) and \( m = -\frac{3}{2} \). Substituting these values:
\[
y - 10 = -\frac{3}{2}(x + 3)
\]
Simplify:
\[
y - 10 = -\frac{3}{2}x - \frac{9}{2}
\]
\[
y = -\frac{3}{2}x - \frac{9}{2} + 10
\]
\[
y = -\frac{3}{2}x - \frac{9}{2} + \frac{20}{2}
\]
\[
y = -\frac{3}{2}x + \frac{11}{2}
\]

Thus, the equation of the line \( RS \) is:
\[
y = -\frac{3}{2}x + \frac{11}{2}
\]

---

### Final Answers:
(a) The equation of the line \( QR \) is:
\[
y = 3
\]

(b) The equation of the line \( RS \) is:
\[
y = -\frac{3}{2}x + \frac{11}{2}
\]

---

Do you have any questions, or would you like further explanations?

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### 5 Related Questions:
1. How do we derive the equation of a line in point-slope form?
2. What is the significance of slope for parallel and perpendicular lines?
3. How do we convert between general form and slope-intercept form of a line?
4. How can you determine if two lines are parallel or perpendicular given their equations?
5. What is the geometric meaning of a line parallel to the x-axis?

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**Tip:** When a line is parallel to the x-axis, all points on the line share the same y-coordinate, making the equation of the line very straightforward.

In Diagram 4, PQ, QR, QS, and RS are straight lines drawn on a Cartesian plane. Given that the equation of PQ is 3x + 2y = 6, state the equation of line QR and determine the equation of line RS. Point S has coordinates (-3, 10), and point Q lies on the y-axis.

Learn how to find the equations of lines QR and RS based on the given Cartesian plane. The problem involves concepts like parallel lines, slopes, and line equations. This is suitable for students in Grades 10-12.

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