Math Problem Statement

A triangle has vertices at A(2, —2), B(—4, —4), and C(O, 4).a) Draw the triangle, and determine the coordinates of the midpoints of its sides. b) Draw the median from vertex A, and determine its equation.

Solution

a) Draw the triangle and determine the coordinates of the midpoints of its sides.

  1. Vertices of the triangle:

    • A(2,2)A(2, -2)
    • B(4,4)B(-4, -4)
    • C(0,4)C(0, 4)

    To find the midpoints of each side, we'll use the midpoint formula. The midpoint MM of two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by:

    M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

  2. Midpoint of side ABAB:

    • Using A(2,2)A(2, -2) and B(4,4)B(-4, -4): MAB=(2+(4)2,2+(4)2)=(22,62)=(1,3)M_{AB} = \left( \frac{2 + (-4)}{2}, \frac{-2 + (-4)}{2} \right) = \left( \frac{-2}{2}, \frac{-6}{2} \right) = (-1, -3)
  3. Midpoint of side BCBC:

    • Using B(4,4)B(-4, -4) and C(0,4)C(0, 4): MBC=(4+02,4+42)=(42,02)=(2,0)M_{BC} = \left( \frac{-4 + 0}{2}, \frac{-4 + 4}{2} \right) = \left( \frac{-4}{2}, \frac{0}{2} \right) = (-2, 0)
  4. Midpoint of side ACAC:

    • Using A(2,2)A(2, -2) and C(0,4)C(0, 4): MAC=(2+02,2+42)=(22,22)=(1,1)M_{AC} = \left( \frac{2 + 0}{2}, \frac{-2 + 4}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)

    Therefore, the midpoints of the sides are:

    • Midpoint of ABAB: (1,3)(-1, -3)
    • Midpoint of BCBC: (2,0)(-2, 0)
    • Midpoint of ACAC: (1,1)(1, 1)

b) Draw the median from vertex AA and determine its equation.

A median of a triangle is a line segment that joins a vertex to the midpoint of the opposite side. We will now determine the equation of the median from vertex A(2,2)A(2, -2) to the midpoint of side BCBC, which we found to be (2,0)(-2, 0).

  1. Find the slope of the median: The slope mm between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by:

    m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}

    Using A(2,2)A(2, -2) and (2,0)(-2, 0):

    m=0(2)22=24=12m = \frac{0 - (-2)}{-2 - 2} = \frac{2}{-4} = -\frac{1}{2}

  2. Equation of the median: The equation of a line in point-slope form is:

    yy1=m(xx1)y - y_1 = m(x - x_1)

    Substituting m=12m = -\frac{1}{2} and (x1,y1)=(2,2)(x_1, y_1) = (2, -2):

    y(2)=12(x2)y - (-2) = -\frac{1}{2}(x - 2) Simplifying: y+2=12x+1y + 2 = -\frac{1}{2}x + 1 y=12x+12y = -\frac{1}{2}x + 1 - 2 y=12x1y = -\frac{1}{2}x - 1

Thus, the equation of the median from vertex AA is: y=12x1y = -\frac{1}{2}x - 1

Summary:

  • The midpoints of the sides are:
    • AB:(1,3)AB: (-1, -3)
    • BC:(2,0)BC: (-2, 0)
    • AC:(1,1)AC: (1, 1)
  • The equation of the median from AA to BCBC is: y=12x1y = -\frac{1}{2}x - 1

Would you like me to provide more details or a diagram for this?

Here are some follow-up questions to deepen your understanding:

  1. What is the significance of the centroid in relation to the medians of a triangle?
  2. How would you find the length of the median from vertex AA to side BCBC?
  3. Can you determine the equation of the other two medians in the triangle?
  4. What are the properties of medians in an equilateral triangle?
  5. How do the midpoints and the centroid of a triangle relate geometrically?

Tip: The point where the medians of a triangle intersect is called the centroid, which divides each median into a 2:1 ratio.

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Math Problem Analysis

Mathematical Concepts

Geometry
Coordinate Geometry
Triangles
Medians

Formulas

Midpoint formula: M = [(x1 + x2)/2, (y1 + y2)/2]
Slope formula: m = (y2 - y1)/(x2 - x1)
Point-slope form equation: y - y1 = m(x - x1)

Theorems

Median of a triangle theorem
Properties of medians

Suitable Grade Level

Grades 9-10