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4.3 Q-12

Question Statement

Find the equations of the sides, altitudes, and medians of a triangle with vertices at A(βˆ’3,2)A(-3, 2), B(5,4)B(5, 4), and C(3,βˆ’8)C(3, -8).


Background and Explanation

To solve this, we need to use the following concepts:

  • Slope of a Line: The slope between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by:
    m=y2βˆ’y1x2βˆ’x1m = \frac{y_2 - y_1}{x_2 - x_1}
  • Midpoint Formula: The midpoint of two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is:
    (x1+x22,y1+y22)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
  • Point-Slope Form of a Line: The equation of a line passing through a point (x1,y1)(x_1, y_1) with slope mm is:
    yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1)
  • Perpendicular Line: The slope of a line perpendicular to another with slope mm is βˆ’1m-\frac{1}{m}.

Solution

Step 1: Finding the Equation of the Sides

1. Side ABβ€Ύ\overline{AB}

  • Slope of ABβ€Ύ\overline{AB}:
    m1=4βˆ’25βˆ’(βˆ’3)=28=14m_1 = \frac{4 - 2}{5 - (-3)} = \frac{2}{8} = \frac{1}{4}
  • Using the point-slope form for point A(βˆ’3,2)A(-3, 2):
    yβˆ’2=14(xβˆ’(βˆ’3))y - 2 = \frac{1}{4}(x - (-3))
    Simplifying:
    4(yβˆ’2)=x+3β‡’4yβˆ’8=x+3β‡’xβˆ’4y+11=04(y - 2) = x + 3 \quad \Rightarrow \quad 4y - 8 = x + 3 \quad \Rightarrow \quad x - 4y + 11 = 0
    So, the equation of side ABβ€Ύ\overline{AB} is:
    xβˆ’4y+11=0x - 4y + 11 = 0

2. Side BCβ€Ύ\overline{BC}

  • Slope of BCβ€Ύ\overline{BC}:
    m2=βˆ’8βˆ’43βˆ’5=βˆ’12βˆ’2=6m_2 = \frac{-8 - 4}{3 - 5} = \frac{-12}{-2} = 6
  • Using the point-slope form for point B(5,4)B(5, 4):
    yβˆ’4=6(xβˆ’5)y - 4 = 6(x - 5)
    Simplifying:
    yβˆ’4=6xβˆ’30β‡’6xβˆ’yβˆ’26=0y - 4 = 6x - 30 \quad \Rightarrow \quad 6x - y - 26 = 0
    So, the equation of side BCβ€Ύ\overline{BC} is:
    6xβˆ’yβˆ’26=06x - y - 26 = 0

3. Side ACβ€Ύ\overline{AC}

  • Slope of ACβ€Ύ\overline{AC}:
    m3=βˆ’8βˆ’23βˆ’(βˆ’3)=βˆ’106=βˆ’53m_3 = \frac{-8 - 2}{3 - (-3)} = \frac{-10}{6} = \frac{-5}{3}
  • Using the point-slope form for point A(βˆ’3,2)A(-3, 2):
    yβˆ’2=βˆ’53(xβˆ’3)y - 2 = -\frac{5}{3}(x - 3)
    Simplifying:
    3(yβˆ’2)=βˆ’5(x+3)β‡’3yβˆ’6=βˆ’5xβˆ’15β‡’5x+3y+9=03(y - 2) = -5(x + 3) \quad \Rightarrow \quad 3y - 6 = -5x - 15 \quad \Rightarrow \quad 5x + 3y + 9 = 0
    So, the equation of side ACβ€Ύ\overline{AC} is:
    5x+3y+9=05x + 3y + 9 = 0

Step 2: Finding the Equations of the Altitudes

1. Altitude through Vertex AA

  • The slope of the perpendicular line to ABβ€Ύ\overline{AB} is βˆ’4-4 (since the slope of ABβ€Ύ\overline{AB} is 14\frac{1}{4}).
  • Using point A(βˆ’3,2)A(-3, 2) and slope βˆ’4-4:
    yβˆ’2=βˆ’4(xβˆ’(βˆ’3))y - 2 = -4(x - (-3))
    Simplifying:
    6(yβˆ’2)=βˆ’1(x+3)β‡’x+6yβˆ’9=06(y - 2) = -1(x + 3) \quad \Rightarrow \quad x + 6y - 9 = 0
    So, the equation of the altitude through vertex AA is:
    x+6yβˆ’9=0x + 6y - 9 = 0

2. Altitude through Vertex BB

  • The slope of the perpendicular line to BCβ€Ύ\overline{BC} is βˆ’16-\frac{1}{6} (since the slope of BCβ€Ύ\overline{BC} is 66).
  • Using point B(5,4)B(5, 4) and slope βˆ’16-\frac{1}{6}:
    yβˆ’4=βˆ’16(xβˆ’5)y - 4 = -\frac{1}{6}(x - 5)
    Simplifying:
    6(yβˆ’4)=βˆ’1(xβˆ’5)β‡’3xβˆ’5y+5=06(y - 4) = -1(x - 5) \quad \Rightarrow \quad 3x - 5y + 5 = 0
    So, the equation of the altitude through vertex BB is:
    3xβˆ’5y+5=03x - 5y + 5 = 0

3. Altitude through Vertex CC

  • The slope of the perpendicular line to ACβ€Ύ\overline{AC} is βˆ’35-\frac{3}{5} (since the slope of ACβ€Ύ\overline{AC} is βˆ’53\frac{-5}{3}).
  • Using point C(3,βˆ’8)C(3, -8) and slope βˆ’35-\frac{3}{5}:
    yβˆ’(βˆ’8)=βˆ’35(xβˆ’3)y - (-8) = -\frac{3}{5}(x - 3)
    Simplifying:
    5(y+8)=βˆ’3(xβˆ’3)β‡’4x+yβˆ’4=05(y + 8) = -3(x - 3) \quad \Rightarrow \quad 4x + y - 4 = 0
    So, the equation of the altitude through vertex CC is:
    4x+yβˆ’4=04x + y - 4 = 0

Step 3: Finding the Equations of the Medians

1. Median through Vertex AA

  • The midpoint of BCβ€Ύ\overline{BC} is (1,3)(1, 3).
  • The slope of the median through AA is the slope between A(βˆ’3,2)A(-3, 2) and (1,3)(1, 3):
    m=3βˆ’21βˆ’(βˆ’3)=14m = \frac{3 - 2}{1 - (-3)} = \frac{1}{4}
  • Using point A(βˆ’3,2)A(-3, 2) and slope βˆ’47\frac{-4}{7}:
    yβˆ’2=βˆ’47(xβˆ’(βˆ’3))y - 2 = -\frac{4}{7}(x - (-3))
    Simplifying:
    4x+7yβˆ’2=04x + 7y - 2 = 0

2. Median through Vertex BB

  • The midpoint of ACβ€Ύ\overline{AC} is (0,βˆ’3)(0, -3).
  • The slope of the median through vertex B is
    βˆ’3βˆ’40βˆ’5=75\frac{-3 - 4}{0 - 5}=\frac{7}{5} The equation of the median:
yβˆ’4=7/5(xβˆ’5)wherearefoundtoo....y-4=7/5(x-5) \text where are found too....