4.4 Q-15

Question Statement

Given the vertices of a triangle as $A(-2, 3)$, $B(-4, 1)$, and $C(3, 5)$, find the center of the circumcircle of the triangle by determining the point of intersection of the perpendicular bisectors of its sides.

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Background and Explanation

In this problem, we are tasked with finding the circumcenter of the triangle, which is the point of intersection of the perpendicular bisectors of the triangle’s sides. The circumcenter is equidistant from all three vertices of the triangle. To find it, we need to:

1. Find the midpoints of two sides of the triangle.
2. Determine the equations of the perpendicular bisectors for these sides.
3. Solve the system of linear equations to find the point of intersection, which gives the circumcenter.

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Solution

Step 1: Calculate the midpoints of sides $BC$ and $AC$.

• Midpoint of $\overline{BC}$:

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

• Midpoint of $\overline{AC}$:

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

Step 2: Calculate the slopes of the sides.

• Slope of $\overline{BC}$:

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

• Slope of $\overline{AC}$:

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

Step 3: Find the equations of the perpendicular bisectors.

• The perpendicular bisector of a line has a slope that is the negative reciprocal of the original line’s slope.

• Perpendicular bisector of $\overline{BC}$:
  The slope is $-\frac{7}{4}$ (negative reciprocal of $\frac{4}{7}$). Using the point $D\left(-\frac{1}{2}, 3\right)$, the equation of the perpendicular bisector is:

$$y - 3 = -\frac{7}{4} \left(x + \frac{1}{2}\right)$$

Simplifying:

$$4(y - 3) = -7 \left(x + \frac{1}{2}\right) \quad \Rightarrow \quad 7x + 4y - 17 = 0$$

(Equation 1)

• Perpendicular bisector of $\overline{AC}$:
  The slope is $-\frac{5}{2}$ (negative reciprocal of $\frac{2}{5}$). Using the point $E\left(\frac{1}{2}, 4\right)$, the equation of the perpendicular bisector is:

$$y - 4 = -\frac{5}{2} \left(x - \frac{1}{2}\right)$$

Simplifying:

$$2(y - 4) = -5 \left(x - \frac{1}{2}\right) \quad \Rightarrow \quad 10x + 4y - 21 = 0$$

(Equation 2)

Step 4: Solve the system of equations.

Now we solve the system of equations:

1. $14x + 8y - 17 = 0$
2. $10x + 4y - 21 = 0$

By solving these two equations simultaneously, we get:

$$x = \frac{25}{6}, \quad y = -\frac{31}{6}$$

Thus, the circumcenter (center of the circumcircle) is the point:

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

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Key Formulas or Methods Used

• Midpoint Formula:

$$\text{Midpoint of } \overline{AB} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$

• Slope Formula:

$$\text{Slope of line } \overline{AB} = \frac{y_2 - y_1}{x_2 - x_1}$$

• Equation of Perpendicular Bisector: If the slope of a line is $m$, the slope of the perpendicular line is $-\frac{1}{m}$. The equation of the perpendicular bisector can be written as:

$$y - y_1 = -\frac{1}{m}(x - x_1)$$

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Summary of Steps

1. Find the midpoints of sides $BC$ and $AC$.
2. Calculate the slopes of sides $BC$ and $AC$.
3. Determine the equations of the perpendicular bisectors using the slopes and midpoints.
4. Solve the system of equations to find the point of intersection, which is the circumcenter.