The point at which the line segment joining A(1, 1) and B (5, 5) subtends an obtuse angle is

Options

A. (7, 7)
B. (0, 5)
C. (2, 4)
D. (1, 5)

Answer

To find the point at which the line segment joining A(1, 1) and B(5, 5) subtends an obtuse angle, we can follow these steps:

### Step 1: Determine the equation of the circle with AB as the diameter.
The endpoints of the diameter are A(1, 1) and B(5, 5). The general equation of a circle with diameter endpoints (x1, y1) and (x2, y2) is given by:

\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]

Substituting the coordinates of A and B:

\[ (x - 1)(x - 5) + (y - 1)(y - 5) = 0 \]

### Step 2: Expand the equation.
Expanding the equation:

\[ (x^2 - 5x - x + 5) + (y^2 - 5y - y + 5) = 0 \]

This simplifies to:

\[ x^2 + y^2 - 6x - 6y + 10 = 0 \]

### Step 3: Identify the condition for obtuse angles.
A point P(h, k) will subtend an obtuse angle at the endpoints A and B if it lies inside the circle defined by the equation derived in Step 2. For a point to be inside the circle, the value of the left-hand side of the circle's equation must be less than 0:

\[ h^2 + k^2 - 6h - 6k + 10 < 0 \]

### Step 4: Test points to find one that satisfies the condition.
We will test various points to find one that satisfies the inequality.

1. **Test Point (7, 7)**:
\[ 7^2 + 7^2 - 6 \cdot 7 - 6 \cdot 7 + 10 = 49 + 49 - 42 - 42 + 10 = 24 \quad (\text{Outside}) \]

2. **Test Point (0, 5)**:
\[ 0^2 + 5^2 - 6 \cdot 0 - 6 \cdot 5 + 10 = 0 + 25 - 0 - 30 + 10 = 5 \quad (\text{Outside}) \]

3. **Test Point (2, 4)**:
\[ 2^2 + 4^2 - 6 \cdot 2 - 6 \cdot 4 + 10 = 4 + 16 - 12 - 24 + 10 = -6 \quad (\text{Inside}) \]

4. **Test Point (1, 5)**:
\[ 1^2 + 5^2 - 6 \cdot 1 - 6 \cdot 5 + 10 = 1 + 25 - 6 - 30 + 10 = 0 \quad (\text{On the circle}) \]

### Conclusion:
The point (2, 4) is inside the circle, and therefore, it is the point at which the line segment AB subtends an obtuse angle.

### Final Answer:
The point at which the line segment joining A(1, 1) and B(5, 5) subtends an obtuse angle is **(2, 4)**.

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