4.3 Q-12

Question Statement

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

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Background and Explanation

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

• Slope of a Line: The slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
  $m = \frac{y_2 - y_1}{x_2 - x_1}$
• Midpoint Formula: The midpoint of two points $(x_1, y_1)$ and $(x_2, y_2)$ is:
  $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
• Point-Slope Form of a Line: The equation of a line passing through a point $(x_1, y_1)$ with slope $m$ is:
  $y - y_1 = m(x - x_1)$
• Perpendicular Line: The slope of a line perpendicular to another with slope $m$ is $-\frac{1}{m}$.

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Solution

Step 1: Finding the Equation of the Sides

1. Side $\overline{AB}$

• Slope of $\overline{AB}$:
  $m_1 = \frac{4 - 2}{5 - (-3)} = \frac{2}{8} = \frac{1}{4}$
• Using the point-slope form for point $A(-3, 2)$:
  $y - 2 = \frac{1}{4}(x - (-3))$
  Simplifying:
  $4(y - 2) = x + 3 \quad \Rightarrow \quad 4y - 8 = x + 3 \quad \Rightarrow \quad x - 4y + 11 = 0$
  So, the equation of side $\overline{AB}$ is:
  $x - 4y + 11 = 0$

2. Side $\overline{BC}$

• Slope of $\overline{BC}$:
  $m_2 = \frac{-8 - 4}{3 - 5} = \frac{-12}{-2} = 6$
• Using the point-slope form for point $B(5, 4)$:
  $y - 4 = 6(x - 5)$
  Simplifying:
  $y - 4 = 6x - 30 \quad \Rightarrow \quad 6x - y - 26 = 0$
  So, the equation of side $\overline{BC}$ is:
  $6x - y - 26 = 0$

3. Side $\overline{AC}$

• Slope of $\overline{AC}$:
  $m_3 = \frac{-8 - 2}{3 - (-3)} = \frac{-10}{6} = \frac{-5}{3}$
• Using the point-slope form for point $A(-3, 2)$:
  $y - 2 = -\frac{5}{3}(x - 3)$
  Simplifying:
  $3(y - 2) = -5(x + 3) \quad \Rightarrow \quad 3y - 6 = -5x - 15 \quad \Rightarrow \quad 5x + 3y + 9 = 0$
  So, the equation of side $\overline{AC}$ is:
  $5x + 3y + 9 = 0$

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Step 2: Finding the Equations of the Altitudes

1. Altitude through Vertex $A$

• The slope of the perpendicular line to $\overline{AB}$ is $-4$ (since the slope of $\overline{AB}$ is $\frac{1}{4}$).
• Using point $A(-3, 2)$ and slope $-4$:
  $y - 2 = -4(x - (-3))$
  Simplifying:
  $6(y - 2) = -1(x + 3) \quad \Rightarrow \quad x + 6y - 9 = 0$
  So, the equation of the altitude through vertex $A$ is:
  $x + 6y - 9 = 0$

2. Altitude through Vertex $B$

• The slope of the perpendicular line to $\overline{BC}$ is $-\frac{1}{6}$ (since the slope of $\overline{BC}$ is $6$).
• Using point $B(5, 4)$ and slope $-\frac{1}{6}$:
  $y - 4 = -\frac{1}{6}(x - 5)$
  Simplifying:
  $6(y - 4) = -1(x - 5) \quad \Rightarrow \quad 3x - 5y + 5 = 0$
  So, the equation of the altitude through vertex $B$ is:
  $3x - 5y + 5 = 0$

3. Altitude through Vertex $C$

• The slope of the perpendicular line to $\overline{AC}$ is $-\frac{3}{5}$ (since the slope of $\overline{AC}$ is $\frac{-5}{3}$).
• Using point $C(3, -8)$ and slope $-\frac{3}{5}$:
  $y - (-8) = -\frac{3}{5}(x - 3)$
  Simplifying:
  $5(y + 8) = -3(x - 3) \quad \Rightarrow \quad 4x + y - 4 = 0$
  So, the equation of the altitude through vertex $C$ is:
  $4x + y - 4 = 0$

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Step 3: Finding the Equations of the Medians

1. Median through Vertex $A$

• The midpoint of $\overline{BC}$ is $(1, 3)$.
• The slope of the median through $A$ is the slope between $A(-3, 2)$ and $(1, 3)$:
  $m = \frac{3 - 2}{1 - (-3)} = \frac{1}{4}$
• Using point $A(-3, 2)$ and slope $\frac{-4}{7}$:
  $y - 2 = -\frac{4}{7}(x - (-3))$
  Simplifying:
  $4x + 7y - 2 = 0$

2. Median through Vertex $B$

• The midpoint of $\overline{AC}$ is $(0, -3)$.
• The slope of the median through vertex B is
  $\frac{-3 - 4}{0 - 5}=\frac{7}{5}$ The equation of the median:

$$y-4=7/5(x-5) \text where are found too....$$