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The problem in the image is asking to find the value of \( k \) for which \( \angle \hat{B} = 90^\circ \), meaning that the triangle has a right angle at point \( B \).

The points of the triangle are:
- \( A(2, 3) \)
- \( B(k, 1) \)
- \( C(-1, 0) \)

To solve this, we need to use the concept that two lines are perpendicular if the product of their slopes is equal to \( -1 \).

### Steps:
1. **Find the slope of line \( AB \)**:
   The slope of a line between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
   \[
   m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}
   \]
   For \( A(2, 3) \) and \( B(k, 1) \), the slope of \( AB \) is:
   \[
   m_{AB} = \frac{1 - 3}{k - 2} = \frac{-2}{k - 2}
   \]

2. **Find the slope of line \( BC \)**:
   Using the points \( B(k, 1) \) and \( C(-1, 0) \), the slope of \( BC \) is:
   \[
   m_{BC} = \frac{1 - 0}{k - (-1)} = \frac{1}{k + 1}
   \]

3. **Set up the perpendicularity condition**:
   Since \( \angle \hat{B} = 90^\circ \), the lines \( AB \) and \( BC \) must be perpendicular, meaning:
   \[
   m_{AB} \cdot m_{BC} = -1
   \]
   Substituting the slopes:
   \[
   \frac{-2}{k - 2} \cdot \frac{1}{k + 1} = -1
   \]

4. **Solve for \( k \)**:
   Now solve the equation:
   \[
   \frac{-2}{(k - 2)(k + 1)} = -1
   \]
   Multiply both sides by \( (k - 2)(k + 1) \):
   \[
   -2 = -(k - 2)(k + 1)
   \]
   Simplify the right-hand side:
   \[
   -2 = -(k^2 - k - 2)
   \]
   Distribute the negative sign:
   \[
   -2 = -k^2 + k + 2
   \]
   Bring all terms to one side:
   \[
   k^2 - k - 4 = 0
   \]
   Now, solve this quadratic equation using the quadratic formula:
   \[
   k = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)} = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}
   \]
   So, the two possible values for \( k \) are:
   \[
   k = \frac{1 + \sqrt{17}}{2} \quad \text{or} \quad k = \frac{1 - \sqrt{17}}{2}
   \]

Would you like more details on the steps or another part of the solution explained?

Here are some related questions for further practice:
1. What is the equation of a line passing through points \( A(2, 3) \) and \( C(-1, 0) \)?
2. How can we verify that the triangle is right-angled using the distance formula?
3. What is the area of the triangle formed by points \( A(2, 3) \), \( B(k, 1) \), and \( C(-1, 0) \)?
4. How does the value of \( k \) affect the orientation of the triangle?
5. If \( \angle \hat{A} = 90^\circ \), how would we find the value of \( k \)?

**Tip**: Always check the perpendicularity condition by verifying the product of the slopes when solving geometry problems involving right angles.

Given points A(2, 3), B(k, 1), and C(-1, 0), find the value of k such that angle ∠B is 90°, i.e., AB ⊥ BC.

Learn how to find the value of k for which a triangle formed by points A(2, 3), B(k, 1), and C(-1, 0) is right-angled at B. This problem involves using slope formulas, the condition for perpendicularity, and solving a quadratic equation. Suitable for students in Grades 9-10.

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