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

Options

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

Answer

To determine the condition under which the medians AD and BE of triangle ABC are mutually perpendicular, we will follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are given as:
- A(0, b)
- B(0, 0)
- C(a, 0)

### Step 2: Find the midpoints D and E
- **Midpoint D** of side BC:
\[ D = \left( \frac{0 + a}{2}, \frac{0 + 0}{2} \right) = \left( \frac{a}{2}, 0 \right) \]

- **Midpoint E** of side AC:
\[ E = \left( \frac{0 + a}{2}, \frac{b + 0}{2} \right) = \left( \frac{a}{2}, \frac{b}{2} \right) \]

### Step 3: Calculate the slopes of the medians AD and BE
- **Slope of median AD**:
\[ \text{slope of } AD = \frac{0 - b}{\frac{a}{2} - 0} = \frac{-b}{\frac{a}{2}} = \frac{-2b}{a} \]

- **Slope of median BE**:
\[ \text{slope of } BE = \frac{\frac{b}{2} - 0}{\frac{a}{2} - 0} = \frac{\frac{b}{2}}{\frac{a}{2}} = \frac{b}{a} \]

### Step 4: Set the product of the slopes equal to -1
For the medians to be perpendicular, the product of their slopes must equal -1:
\[ \left( \frac{-2b}{a} \right) \left( \frac{b}{a} \right) = -1 \]

### Step 5: Simplify the equation
\[ \frac{-2b^2}{a^2} = -1 \]
Multiplying both sides by -1 gives:
\[ \frac{2b^2}{a^2} = 1 \]

### Step 6: Rearranging the equation
This can be rearranged to:
\[ 2b^2 = a^2 \]
Taking the square root of both sides yields:
\[ a = \pm \sqrt{2}b \]

### Final Result
Thus, the medians AD and BE of triangle ABC are mutually perpendicular if:
\[ a = \pm b\sqrt{2} \]