The vertices of a triangle ABC are (2,1), (5,2) and (3,4) respectively. The circumcentre is the point

Options

A. `(3/4,7/4)`
B. `(13/4,9/4)`
C. `(11/4,5/4)`
D. None

Answer

To find the circumcenter of triangle ABC with vertices A(2, 1), B(5, 2), and C(3, 4), we will follow these steps:

### Step 1: Understand the Circumcenter
The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect. It is equidistant from all three vertices of the triangle.

### Step 2: Find the Midpoints of Two Sides
Let's find the midpoints of sides AB and AC.

- Midpoint of AB:
\[ M_{AB} = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right) = \left( \frac{2 + 5}{2}, \frac{1 + 2}{2} \right) = \left( \frac{7}{2}, \frac{3}{2} \right) \]

- Midpoint of AC:
\[ M_{AC} = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = \left( \frac{2 + 3}{2}, \frac{1 + 4}{2} \right) = \left( \frac{5}{2}, \frac{5}{2} \right) \]

### Step 3: Find the Slopes of AB and AC
Next, we calculate the slopes of lines AB and AC.

- Slope of AB:
\[ m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{2 - 1}{5 - 2} = \frac{1}{3} \]

- Slope of AC:
\[ m_{AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{4 - 1}{3 - 2} = 3 \]

### Step 4: Find the Slopes of the Perpendicular Bisectors
The slopes of the perpendicular bisectors will be the negative reciprocals of the slopes of the sides.

- Slope of the perpendicular bisector of AB:
\[ m_{PB_{AB}} = -\frac{1}{m_{AB}} = -3 \]

- Slope of the perpendicular bisector of AC:
\[ m_{PB_{AC}} = -\frac{1}{m_{AC}} = -\frac{1}{3} \]

### Step 5: Write the Equations of the Perpendicular Bisectors
Using point-slope form \(y - y_1 = m(x - x_1)\):

- Equation of the perpendicular bisector of AB:
\[ y - \frac{3}{2} = -3\left(x - \frac{7}{2}\right) \]
Simplifying:
\[ y - \frac{3}{2} = -3x + \frac{21}{2} \implies y = -3x + 12 \]

- Equation of the perpendicular bisector of AC:
\[ y - \frac{5}{2} = -\frac{1}{3}\left(x - \frac{5}{2}\right) \]
Simplifying:
\[ y - \frac{5}{2} = -\frac{1}{3}x + \frac{5}{6} \implies y = -\frac{1}{3}x + \frac{25}{6} \]

### Step 6: Solve the System of Equations
Now we have two equations:
1. \(y = -3x + 12\)
2. \(y = -\frac{1}{3}x + \frac{25}{6}\)

Setting them equal to find x:
\[ -3x + 12 = -\frac{1}{3}x + \frac{25}{6} \]
Multiplying through by 6 to eliminate fractions:
\[ -18x + 72 = -2x + 25 \]
Rearranging gives:
\[ -16x = -47 \implies x = \frac{47}{16} \]

Substituting \(x\) back to find \(y\):
\[ y = -3\left(\frac{47}{16}\right) + 12 = -\frac{141}{16} + \frac{192}{16} = \frac{51}{16} \]

### Step 7: Conclusion
Thus, the circumcenter of triangle ABC is at the point:
\[ \left(\frac{47}{16}, \frac{51}{16}\right) \]