Math Problem Statement

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.

Solution

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

Given:

Vertices of the triangle PQR:

  • P(3,5)P(-3, 5)
  • Q(7,3)Q(7, 3)
  • R(1,5)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=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) For P(3,5)P(-3, 5) and Q(7,3)Q(7, 3): M=(3+72,5+32)=(42,82)=(2,4)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)R(-1, -5) and point M(2,4)M(2, 4) lie on the median. We can find the equation of the line using the point-slope form: yy1=m(xx1)y - y_1 = m(x - x_1) where mm is the slope of the line, and (x1,y1)(x_1, y_1) is a point on the line. The slope mm between R(1,5)R(-1, -5) and M(2,4)M(2, 4) is: m=4(5)2(1)=93=3m = \frac{4 - (-5)}{2 - (-1)} = \frac{9}{3} = 3 Using point R(1,5)R(-1, -5): y(5)=3(x(1))    y+5=3(x+1)y - (-5) = 3(x - (-1)) \implies y + 5 = 3(x + 1) Expanding: y+5=3x+3    y=3x2y + 5 = 3x + 3 \implies y = 3x - 2 So, the equation of the median RMRM is: y=3x2y = 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 QRQR can be calculated as: mQR=3(5)7(1)=88=1m_{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 QRQR is 1, the slope of the perpendicular line is: mAP=1m_{AP} = -1

  3. Equation of AP: Using the point-slope form again with point P(3,5)P(-3, 5) and slope 1-1: y5=1(x(3))    y5=(x+3)y - 5 = -1(x - (-3)) \implies y - 5 = -(x + 3) Expanding: y5=x3    y=x+2y - 5 = -x - 3 \implies y = -x + 2 So, the equation of the altitude APAP is: y=x+2y = -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=3x2y = 3x - 2 (from the median RM)
  2. y=x+2y = -x + 2 (from the altitude AP)

Set the equations equal to each other: 3x2=x+23x - 2 = -x + 2 Solving for xx: 3x+x=2+2    4x=4    x=13x + x = 2 + 2 \implies 4x = 4 \implies x = 1

Now, substitute x=1x = 1 into either equation to find yy. Using y=3x2y = 3x - 2: y=3(1)2=32=1y = 3(1) - 2 = 3 - 2 = 1

Thus, the point of intersection is: (1,1)(1, 1)

Summary of Answers:

(a) Equation of the median RMRM: y=3x2y = 3x - 2

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

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

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Equations of Lines
Triangles
Medians
Altitudes

Formulas

Midpoint formula: M = ((x1 + x2)/2, (y1 + y2)/2)
Slope formula: m = (y2 - y1) / (x2 - x1)
Point-slope form of a line: y - y1 = m(x - x1)

Theorems

Basic Theorems of Coordinate Geometry
Properties of Medians and Altitudes in a Triangle

Suitable Grade Level

Grades 9-12