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Let's go through each part of the problem systematically.

### Given:
Vertices of the triangle PQR:
- \( P(-3, 5) \)
- \( Q(7, 3) \)
- \( R(-1, -5) \)

### (a) Find the equation of the median RM

A median of a triangle is a line segment that joins a vertex to the midpoint of the opposite side.

1. **Midpoint M of PQ**:
   The midpoint formula is:
   \[
   M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
   \]
   For \( P(-3, 5) \) and \( Q(7, 3) \):
   \[
   M = \left( \frac{-3 + 7}{2}, \frac{5 + 3}{2} \right) = \left( \frac{4}{2}, \frac{8}{2} \right) = (2, 4)
   \]

2. **Equation of RM**:
   Now, we know that point \( R(-1, -5) \) and point \( M(2, 4) \) lie on the median. We can find the equation of the line using the point-slope form:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( m \) is the slope of the line, and \( (x_1, y_1) \) is a point on the line. The slope \( m \) between \( R(-1, -5) \) and \( M(2, 4) \) is:
   \[
   m = \frac{4 - (-5)}{2 - (-1)} = \frac{9}{3} = 3
   \]
   Using point \( R(-1, -5) \):
   \[
   y - (-5) = 3(x - (-1)) \implies y + 5 = 3(x + 1)
   \]
   Expanding:
   \[
   y + 5 = 3x + 3 \implies y = 3x - 2
   \]
   So, the equation of the median \( RM \) is:
   \[
   y = 3x - 2
   \]

### (b) Find the equation of the altitude AP

An altitude of a triangle is a line segment from a vertex perpendicular to the opposite side.

1. **Slope of side QR**:
   The slope of \( QR \) can be calculated as:
   \[
   m_{QR} = \frac{3 - (-5)}{7 - (-1)} = \frac{8}{8} = 1
   \]

2. **Slope of the perpendicular line (altitude AP)**:
   The slope of a line perpendicular to another line is the negative reciprocal of the slope of the other line. Since the slope of \( QR \) is 1, the slope of the perpendicular line is:
   \[
   m_{AP} = -1
   \]

3. **Equation of AP**:
   Using the point-slope form again with point \( P(-3, 5) \) and slope \( -1 \):
   \[
   y - 5 = -1(x - (-3)) \implies y - 5 = -(x + 3)
   \]
   Expanding:
   \[
   y - 5 = -x - 3 \implies y = -x + 2
   \]
   So, the equation of the altitude \( AP \) is:
   \[
   y = -x + 2
   \]

### (c) Find the coordinates of the point of intersection of RM and AP

To find the point of intersection of the two lines, solve the system of equations:
1. \( y = 3x - 2 \) (from the median RM)
2. \( y = -x + 2 \) (from the altitude AP)

Set the equations equal to each other:
\[
3x - 2 = -x + 2
\]
Solving for \( x \):
\[
3x + x = 2 + 2 \implies 4x = 4 \implies x = 1
\]

Now, substitute \( x = 1 \) into either equation to find \( y \). Using \( y = 3x - 2 \):
\[
y = 3(1) - 2 = 3 - 2 = 1
\]

Thus, the point of intersection is:
\[
(1, 1)
\]

### Summary of Answers:
(a) Equation of the median \( RM \): \( y = 3x - 2 \)

(b) Equation of the altitude \( AP \): \( y = -x + 2 \)

(c) Point of intersection of RM and AP: \( (1, 1) \)

Would you like any further clarification or more details? Here are five related questions:

1. How can you find the centroid of a triangle?
2. What is the formula to calculate the area of a triangle given its vertices?
3. Can you determine the length of the median in this triangle?
4. How do you find the orthocenter of a triangle?
5. What properties does the intersection of medians have in a triangle?

**Tip**: In geometry, intersections of special lines like medians and altitudes often give key points (centroid, orthocenter), useful in various applications of triangle properties!

Triangle PQR has vertices P(-3, 5), Q(7, 3), and R(-1, -5). (a) Find the equation of the median RM. (b) Find the equation of the altitude AP. (c) Find the coordinates of the point of intersection of RM and AP.

Learn how to find the equation of the median RM and altitude AP of triangle PQR with vertices P(-3, 5), Q(7, 3), and R(-1, -5). Includes the midpoint and slope formulas, along with the step-by-step solution.

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