Math Problem Statement

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

Solution

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

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

1.1. Midpoint of ABAB:

  • The midpoint MABM_{AB} of side ABAB is: MAB=(1+02,1+(4)2)=(12,32)M_{AB} = \left( \frac{-1 + 0}{2}, \frac{1 + (-4)}{2} \right) = \left( -\frac{1}{2}, -\frac{3}{2} \right)

1.2. Slope of ABAB:

  • The slope of line ABAB is: slope of AB=410(1)=51=5\text{slope of } AB = \frac{-4 - 1}{0 - (-1)} = \frac{-5}{1} = -5

1.3. Slope of the perpendicular bisector of ABAB:

  • The perpendicular bisector's slope is the negative reciprocal of the slope of ABAB: slope of perpendicular bisector of AB=15\text{slope of perpendicular bisector of } AB = \frac{1}{5}

1.4. Equation of the perpendicular bisector of ABAB:

  • Using the point-slope form of a line equation, yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is the point: y+32=15(x+12)y + \frac{3}{2} = \frac{1}{5}\left( x + \frac{1}{2} \right)
  • Simplify: 5y+15=x+125y + 15 = x + \frac{1}{2} x5y=292(Equation 1)x - 5y = \frac{29}{2} \quad \text{(Equation 1)}

2.1. Midpoint of BCBC:

  • The midpoint MBCM_{BC} of side BCBC is: MBC=(0+(1)2,4+(5)2)=(12,92)M_{BC} = \left( \frac{0 + (-1)}{2}, \frac{-4 + (-5)}{2} \right) = \left( -\frac{1}{2}, -\frac{9}{2} \right)

2.2. Slope of BCBC:

  • The slope of line BCBC is: slope of BC=5(4)10=11=1\text{slope of } BC = \frac{-5 - (-4)}{-1 - 0} = \frac{-1}{-1} = 1

2.3. Slope of the perpendicular bisector of BCBC:

  • The perpendicular bisector's slope is the negative reciprocal of the slope of BCBC: slope of perpendicular bisector of BC=1\text{slope of perpendicular bisector of } BC = -1

2.4. Equation of the perpendicular bisector of BCBC:

y+92=1(x+12)y + \frac{9}{2} = -1 \left( x + \frac{1}{2} \right) y=x4(Equation 2)y = -x - 4 \quad \text{(Equation 2)}

Step 2: Solve the two equations simultaneously to find the circumcenter.

  • From Equation 2: y=x4y = -x - 4
  • Substitute into Equation 1: x5(x4)=292x - 5(-x - 4) = \frac{29}{2} x+5x+20=292x + 5x + 20 = \frac{29}{2} 6x+20=2926x + 20 = \frac{29}{2}
  • Multiply everything by 2 to clear the fraction: 12x+40=2912x + 40 = 29 12x=1112x = -11 x=1112x = -\frac{11}{12}
  • Substitute x=1112x = -\frac{11}{12} into y=x4y = -x - 4: y=11124=11124812=3712y = \frac{11}{12} - 4 = \frac{11}{12} - \frac{48}{12} = -\frac{37}{12}

Final Answer:

The coordinates of the circumcenter are (1112,3712)\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.

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Math Problem Analysis

Mathematical Concepts

Geometry
Triangles
Circumcenter

Formulas

Midpoint formula
Slope formula
Perpendicular bisector equation

Theorems

Properties of triangles
Circumcenter properties

Suitable Grade Level

High School