Vertics of a triangle ABC are the points (0,0),(a,0) and `(a/2,(asqrt(3))/2)`. Its incentre is the point

Options

A. `((3a)/4,(sqrt(3)a)/4)`
B. `(a/2,(asqrt(3))/6)`
C. `(a/6,(asqrt(3))/2)`
D. `(a/3,(asqrt(3))/2)`

Answer

To find the incenter of triangle ABC with vertices at points \( A(0,0) \), \( B(a,0) \), and \( C\left(\frac{a}{2}, \frac{a\sqrt{3}}{2}\right) \), we can follow these steps:

### Step 1: Determine the lengths of the sides of the triangle

1. **Calculate the lengths of sides AB, BC, and AC:**
- Length of \( AB \):
\[ AB = \sqrt{(a - 0)^2 + (0 - 0)^2} = \sqrt{a^2} = a \]
- Length of \( BC \):
\[ BC = \sqrt{\left(\frac{a}{2} - a\right)^2 + \left(\frac{a\sqrt{3}}{2} - 0\right)^2} = \sqrt{\left(-\frac{a}{2}\right)^2 + \left(\frac{a\sqrt{3}}{2}\right)^2} \]
\[ = \sqrt{\frac{a^2}{4} + \frac{3a^2}{4}} = \sqrt{a^2} = a \]
- Length of \( AC \):
\[ AC = \sqrt{\left(\frac{a}{2} - 0\right)^2 + \left(\frac{a\sqrt{3}}{2} - 0\right)^2} = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a\sqrt{3}}{2}\right)^2} \]
\[ = \sqrt{\frac{a^2}{4} + \frac{3a^2}{4}} = \sqrt{a^2} = a \]

### Step 2: Identify the sides of the triangle

From the calculations, we find that:
- \( AB = a \)
- \( BC = a \)
- \( AC = a \)

Thus, triangle ABC is an equilateral triangle with all sides equal to \( a \).

### Step 3: Use the formula for the incenter

The coordinates of the incenter \( I(x,y) \) of a triangle can be calculated using the formula:
\[ I_x = \frac{a_1x_1 + a_2x_2 + a_3x_3}{a_1 + a_2 + a_3} \]
\[ I_y = \frac{a_1y_1 + a_2y_2 + a_3y_3}{a_1 + a_2 + a_3} \]
where \( a_1, a_2, a_3 \) are the lengths of the sides opposite to vertices \( A, B, C \) respectively, and \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) are the coordinates of the vertices.

### Step 4: Substitute the values into the formula

For triangle ABC:
- \( a_1 = BC = a \)
- \( a_2 = AC = a \)
- \( a_3 = AB = a \)

Coordinates of vertices:
- \( A(0,0) \)
- \( B(a,0) \)
- \( C\left(\frac{a}{2}, \frac{a\sqrt{3}}{2}\right) \)

Substituting into the formula:
\[ I_x = \frac{a \cdot 0 + a \cdot a + a \cdot \frac{a}{2}}{a + a + a} = \frac{0 + a^2 + \frac{a^2}{2}}{3a} = \frac{\frac{3a^2}{2}}{3a} = \frac{a}{2} \]
\[ I_y = \frac{a \cdot 0 + a \cdot 0 + a \cdot \frac{a\sqrt{3}}{2}}{a + a + a} = \frac{0 + 0 + \frac{a^2\sqrt{3}}{2}}{3a} = \frac{\frac{a^2\sqrt{3}}{2}}{3a} = \frac{a\sqrt{3}}{6} \]

### Step 5: Conclusion

Thus, the coordinates of the incenter \( I \) are:
\[ I\left(\frac{a}{2}, \frac{a\sqrt{3}}{6}\right) \]