Find the coordinates of the circumcentre of AABC with vertices A (-1, 1), B (0, - 4) and C(-1, - 5).

To find the coordinates of the circumcenter of triangle \( \triangle ABC \) with vertices \( A(-1, 1) \), \( B(0, -4) \), and \( C(-1, -5) \), follow these steps:

### Step 1: Determine the perpendicular bisectors of two sides of the triangle.

#### 1.1. Midpoint of \( AB \):
- The midpoint \( M_{AB} \) of side \( AB \) is:
\[
M_{AB} = \left( \frac{-1 + 0}{2}, \frac{1 + (-4)}{2} \right) = \left( -\frac{1}{2}, -\frac{3}{2} \right)
\]

#### 1.2. Slope of \( AB \):
- The slope of line \( AB \) is:
\[
\text{slope of } AB = \frac{-4 - 1}{0 - (-1)} = \frac{-5}{1} = -5
\]

#### 1.3. Slope of the perpendicular bisector of \( AB \):
- The perpendicular bisector's slope is the negative reciprocal of the slope of \( AB \):
\[
\text{slope of perpendicular bisector of } AB = \frac{1}{5}
\]

#### 1.4. Equation of the perpendicular bisector of \( AB \):
- Using the point-slope form of a line equation, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point:
\[
y + \frac{3}{2} = \frac{1}{5}\left( x + \frac{1}{2} \right)
\]
- Simplify:
\[
5y + 15 = x + \frac{1}{2}
\]
\[
x - 5y = \frac{29}{2} \quad \text{(Equation 1)}
\]

#### 2.1. Midpoint of \( BC \):
- The midpoint \( M_{BC} \) of side \( BC \) is:
\[
M_{BC} = \left( \frac{0 + (-1)}{2}, \frac{-4 + (-5)}{2} \right) = \left( -\frac{1}{2}, -\frac{9}{2} \right)
\]

#### 2.2. Slope of \( BC \):
- The slope of line \( BC \) is:
\[
\text{slope of } BC = \frac{-5 - (-4)}{-1 - 0} = \frac{-1}{-1} = 1
\]

#### 2.3. Slope of the perpendicular bisector of \( BC \):
- The perpendicular bisector's slope is the negative reciprocal of the slope of \( BC \):
\[
\text{slope of perpendicular bisector of } BC = -1
\]

#### 2.4. Equation of the perpendicular bisector of \( BC \):
\[
y + \frac{9}{2} = -1 \left( x + \frac{1}{2} \right)
\]
\[
y = -x - 4 \quad \text{(Equation 2)}
\]

### Step 2: Solve the two equations simultaneously to find the circumcenter.
- From Equation 2:
\[
y = -x - 4
\]
- Substitute into Equation 1:
\[
x - 5(-x - 4) = \frac{29}{2}
\]
\[
x + 5x + 20 = \frac{29}{2}
\]
\[
6x + 20 = \frac{29}{2}
\]
- Multiply everything by 2 to clear the fraction:
\[
12x + 40 = 29
\]
\[
12x = -11
\]
\[
x = -\frac{11}{12}
\]
- Substitute \( x = -\frac{11}{12} \) into \( y = -x - 4 \):
\[
y = \frac{11}{12} - 4 = \frac{11}{12} - \frac{48}{12} = -\frac{37}{12}
\]

### Final Answer:
The coordinates of the circumcenter are \( \left( -\frac{11}{12}, -\frac{37}{12} \right) \).

---

Would you like more details on any of these steps or have any questions?

Here are 5 related questions:
1. How do you find the circumcenter for a right-angled triangle?
2. What is the significance of the circumcenter in a triangle?
3. How does the circumcenter relate to the triangle's circumcircle?
4. What are the properties of the circumcenter in different types of triangles (acute, obtuse)?
5. How would the circumcenter calculation differ in 3D space?

**Tip:** The circumcenter is equidistant from all three vertices of a triangle, making it the center of the circumcircle.

Learn how to find the coordinates of the circumcenter of triangle ABC with vertices A(-1, 1), B(0, -4), and C(-1, -5). This detailed solution includes steps for calculating perpendicular bisectors, solving equations, and finding geometric centers. Suitable for high school geometry students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation