The points A(3,0) , B (a, -2) and C(4, -1) are the vertices of triangle ABC right angled at vertex A. Find the value of a.

Answer

To find the value of \( a \) such that the points \( A(3,0) \), \( B(a, -2) \), and \( C(4, -1) \) form a right-angled triangle at vertex \( A \), we can use the Pythagorean theorem. The theorem states that for a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

### Step-by-Step Solution:

1. **Identify the points:**
- \( A(3, 0) \)
- \( B(a, -2) \)
- \( C(4, -1) \)

2. **Use the distance formula to find the lengths of the sides:**
The distance formula is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

- **Calculate \( BC \):**
\[ BC = \sqrt{(a - 4)^2 + (-2 + 1)^2} = \sqrt{(a - 4)^2 + (-1)^2} = \sqrt{(a - 4)^2 + 1} \]

- **Calculate \( AB \):**
\[ AB = \sqrt{(3 - a)^2 + (0 + 2)^2} = \sqrt{(3 - a)^2 + 2^2} = \sqrt{(3 - a)^2 + 4} \]

- **Calculate \( AC \):**
\[ AC = \sqrt{(3 - 4)^2 + (0 + 1)^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

3. **Apply the Pythagorean theorem:**
Since triangle \( ABC \) is right-angled at \( A \), we have:
\[ BC^2 = AB^2 + AC^2 \]

Substituting the distances we calculated:
\[ (a - 4)^2 + 1 = (3 - a)^2 + 4 + 2 \]

4. **Simplify the equation:**
\[ (a - 4)^2 + 1 = (3 - a)^2 + 6 \]
Expanding both sides:
\[ (a^2 - 8a + 16) + 1 = (9 - 6a + a^2) + 6 \]
Simplifying further:
\[ a^2 - 8a + 17 = a^2 - 6a + 15 \]

5. **Eliminate \( a^2 \) from both sides:**
\[ -8a + 17 = -6a + 15 \]

6. **Rearranging the equation:**
\[ -8a + 6a = 15 - 17 \]
\[ -2a = -2 \]
Dividing both sides by -2:
\[ a = 1 \]

### Final Answer:
The value of \( a \) is \( 1 \).