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Let's go over the questions and work through them step-by-step:

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### Question 6a:

We are given a right-angled triangle \( PQR \) where:

- \( PQ = 5 \) units
- \( QR = 15 \) units
- Coordinates of point \( P = (-7, 2) \)
- Coordinates of point \( Q = (x, 2) \)

We need to find the coordinates of point \( R \).

Since \( PQ \) is a horizontal line, both points \( P \) and \( Q \) lie on the same horizontal line, i.e., the same y-coordinate. So, \( Q = (x, 2) \). The distance between \( P \) and \( Q \) is given as 5 units.

Using the distance formula:
\[
PQ = |x - (-7)| = 5
\]
\[
|x + 7| = 5 \quad \Rightarrow \quad x + 7 = 5 \quad \text{or} \quad x + 7 = -5
\]
Solving these:
\[
x = -2 \quad \text{or} \quad x = -12
\]

Next, \( QR = 15 \) units and this is a vertical line. Hence, the x-coordinate of \( R \) will be the same as that of \( Q \). So the x-coordinate of \( R \) is either \( -2 \) or \( -12 \).

Thus, \( R = (-2, y) \) or \( R = (-12, y) \). Using the distance between \( Q \) and \( R \) (vertical), we calculate the possible y-coordinates.

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### Question 6b:

In this part, we have a triangle \( ABC \) plotted on a Cartesian plane with:

- Point \( A = (7, -9) \)
- Point \( O \) is the origin \( (0, 0) \)
- Point \( M = (0, n) \) is the midpoint of segment \( BC \)

We are tasked with finding:
1. The coordinates of point \( B \)
2. The value of \( n \)

We'll break down the solution to these parts.

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Would you like me to complete the step-by-step solution for both parts now?

Part a: Find the coordinates of point R in a right-angled triangle PQR using the given distances and coordinates of points P and Q. Part b: Find the coordinates of point B and the value of n for triangle ABC on a Cartesian plane.

This problem involves applying the distance and midpoint formulas to calculate the coordinates of point R in a right-angled triangle PQR and point B in triangle ABC on a Cartesian plane. Suitable for students in Grades 9-11.

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