The medians AD and BE of the triangle ABC with vertices A(0, b), B(0, 0) and C(a, 0) are mutually perpendicular if

Options

A. (a) `b = a sqrt(2)`
B. (b) `a = b sqrt(2)`
C. (c) `b = - a sqrt(2)`
D. (d) `a = -b sqrt(2)`

Answer

To solve the problem, we need to find the condition under which the medians AD and BE of triangle ABC are mutually perpendicular. The vertices of the triangle are given as A(0, b), B(0, 0), and C(a, 0).

### Step-by-step Solution:

1. **Identify the Midpoints of the Sides:**
- The midpoint D of side BC can be calculated using the midpoint formula:
\[ D = \left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}\right) = \left(\frac{0 + a}{2}, \frac{0 + 0}{2}\right) = \left(\frac{a}{2}, 0\right) \]
- The midpoint E of side AC can be calculated similarly:
\[ E = \left(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}\right) = \left(\frac{0 + a}{2}, \frac{b + 0}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right) \]

2. **Find the Slopes of the Medians:**
- The slope of median AD (from A to D) is given by:
\[ m_1 = \frac{y_A - y_D}{x_A - x_D} = \frac{b - 0}{0 - \frac{a}{2}} = \frac{b}{-\frac{a}{2}} = -\frac{2b}{a} \]
- The slope of median BE (from B to E) is given by:
\[ m_2 = \frac{y_B - y_E}{x_B - x_E} = \frac{0 - \frac{b}{2}}{0 - \frac{a}{2}} = \frac{-\frac{b}{2}}{-\frac{a}{2}} = \frac{b}{a} \]

3. **Condition for Perpendicularity:**
- For the two lines to be perpendicular, the product of their slopes must equal -1:
\[ m_1 \cdot m_2 = -1 \]
- Substituting the slopes we found:
\[ \left(-\frac{2b}{a}\right) \cdot \left(\frac{b}{a}\right) = -1 \]
- Simplifying this gives:
\[ -\frac{2b^2}{a^2} = -1 \]
- Removing the negative signs:
\[ \frac{2b^2}{a^2} = 1 \]
- Cross-multiplying yields:
\[ 2b^2 = a^2 \]

4. **Taking Square Roots:**
- Taking the square root of both sides, we find:
\[ a = \pm b\sqrt{2} \]

### Conclusion:
The medians AD and BE of triangle ABC are mutually perpendicular if:
- \( a = b\sqrt{2} \) or \( a = -b\sqrt{2} \)

### Final Answer:
Thus, the correct options are:
- Option B: \( a = b\sqrt{2} \)
- Option D: \( a = -b\sqrt{2} \)