4.4 Q-7

Question Statement

We are given a triangle with vertices at $A(-2,3)$, $B(-4,1)$, and $C(3,5)$. The task is to find the following:

i. The centroid of the triangle.
ii. The orthocenter of the triangle.
iii. The circumcenter of the triangle.

Additionally, we need to determine whether these three points are collinear.

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Background and Explanation

To solve this problem, we need to recall the following concepts:

• Centroid: The point where the three medians of a triangle intersect. It divides each median in a 2:1 ratio.
• Orthocenter: The point where the altitudes of the triangle intersect.
• Circumcenter: The point where the perpendicular bisectors of the sides of the triangle meet.

We will use these definitions to calculate the required points and then check if they are collinear.

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Solution

i. Finding the Centroid

The centroid is the point where all the medians of the triangle intersect. The medians are lines connecting a vertex to the midpoint of the opposite side.

1. Midpoint of BC:
   The midpoint of line segment $\overline{BC}$ is:

$$D = \left( \frac{-4 + 3}{2}, \frac{1 + 5}{2} \right) = \left( -\frac{1}{2}, 3 \right)$$

2. Midpoint of AC:
   The midpoint of line segment $\overline{AC}$ is:

$$E = \left( \frac{-2 + 3}{2}, \frac{3 + 5}{2} \right) = \left( \frac{1}{2}, 4 \right)$$

3. Equation of Median AD:
   The equation of the median from $A$ to $D$ is:

$$\frac{y - 3}{3 - 3} = \frac{x - (-2)}{-\frac{1}{2} - (-2)} \quad \Rightarrow \quad y = 3$$

The centroid is therefore at:

$$(-1, 3)$$

ii. Finding the Orthocenter

The orthocenter is the point of concurrency of the altitudes of the triangle. We need to calculate the equations of two altitudes.

1. Slope of BC:
   The slope of line $\overline{BC}$ is:

$$\text{Slope of } \overline{BC} = \frac{5 - 1}{3 - (-4)} = \frac{4}{7}$$

2. Slope of altitude from A:
   The slope of the altitude from $A$ is the negative reciprocal of the slope of $\overline{BC}$:

$$\text{Slope of altitude from A} = -\frac{7}{4}$$

3. Equation of altitude from A:
   Using the point-slope form of the equation, the equation of the altitude from $A$ is:

$$y - 3 = -\frac{7}{4}(x + 2) \quad \Rightarrow \quad 7x + 4y + 2 = 0$$

4. Slope of AC:
   The slope of line $\overline{AC}$ is:

$$\text{Slope of } \overline{AC} = \frac{5 - 3}{3 - (-2)} = \frac{2}{5}$$

5. Slope of altitude from B:
   The slope of the altitude from $B$ is the negative reciprocal of the slope of $\overline{AC}$:

$$\text{Slope of altitude from B} = \frac{5}{2}$$

6. Equation of altitude from B:
   Using the point-slope form again, the equation of the altitude from $B$ is:

$$y - 1 = -\frac{5}{2}(x + 4) \quad \Rightarrow \quad 5x + 2y + 18 = 0$$

7. Solving the equations of the two altitudes:
   Solving $7x + 4y + 2 = 0$ and $5x + 2y + 18 = 0$ simultaneously, we get:

$$\left[ -\frac{34}{3}, -\frac{58}{3} \right]$$

The orthocenter is at:

$$\left( -\frac{34}{3}, -\frac{58}{3} \right)$$

iii. Finding the Circumcenter

The circumcenter is the point of concurrency of the perpendicular bisectors of the sides of the triangle.

1. Midpoint of BC:
   The midpoint of line $\overline{BC}$ is:

$$P = \left( -\frac{1}{2}, 3 \right)$$

2. Slope of BC:
   The slope of $\overline{BC}$ is:

$$\text{Slope of } \overline{BC} = \frac{4}{7}$$

3. Slope of the perpendicular bisector:
   The slope of the perpendicular bisector of $\overline{BC}$ is:

$$\text{Slope of perpendicular bisector} = -\frac{7}{4}$$

4. Equation of the perpendicular bisector of BC:
   Using the point-slope form, we get the equation of the perpendicular bisector of $\overline{BC}$:

$$7x + 4y - \frac{17}{2} = 0$$

5. Midpoint of AC:
   The midpoint of line $\overline{AC}$ is:

$$P = \left( \frac{1}{2}, 4 \right)$$

6. Slope of AC:
   The slope of $\overline{AC}$ is:

$$\text{Slope of } \overline{AC} = \frac{2}{5}$$

7. Slope of the perpendicular bisector:
   The slope of the perpendicular bisector of $\overline{AC}$ is:

$$\text{Slope of perpendicular bisector} = \frac{5}{2}$$

8. Equation of the perpendicular bisector of AC:
   The equation of the perpendicular bisector of $\overline{AC}$ is:

$$10x + 4y - 21 = 0$$

9. Solving the equations:
   Solving $7x + 4y - \frac{17}{2} = 0$ and $10x + 4y - 21 = 0$, we get:

$$\left( \frac{25}{6}, \frac{-31}{6} \right)$$

The circumcenter is at:

$$\left( \frac{25}{6}, \frac{-31}{6} \right)$$

iv. Are the Centroid, Orthocenter, and Circumcenter Collinear?

To check if the three points are collinear, we calculate the determinant of the matrix formed by their coordinates:

$$\left|\begin{array}{ccc}-1 & 3 & 1 \frac{-34}{3} & \frac{58}{3} & 1 \frac{25}{6} & -\frac{31}{6} & 1\end{array}\right|$$

The calculation results in 0, which means that the three points are indeed collinear.

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Key Formulas or Methods Used

• Centroid: The intersection of the medians of the triangle.
• Orthocenter: The intersection of the altitudes of the triangle.
• Circumcenter: The intersection of the perpendicular bisectors of the sides of the triangle.

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Summary of Steps

1. Find the midpoints of the sides of the triangle.
2. Calculate the equations of the medians and solve to find the centroid.
3. Calculate the equations of the altitudes and solve to find the orthocenter.
4. Calculate the equations of the perpendicular bisectors and solve to find the circumcenter.
5. Check if the centroid, orthocenter, and circumcenter are collinear by calculating the determinant of their coordinates.