The base BC of a triangle ABC is bisected at the point (a, b) and equation to the sides AB and AC are respectively `ax+by=1` and `bx+ay=1` Equation of the median through A is:

Options

A. `ax-by=ab`
B. `(2b-1)(ax+by)=ab`
C. `(2ab-1)(ax+by-1)=(a^(2)+b^(2)-1)(bx+ay-1)`
D. `bx-ay=1`

Answer

To find the equation of the median through point A in triangle ABC, we start with the given information:

1. The base BC of triangle ABC is bisected at the point (a, b).
2. The equations of the sides AB and AC are given as:
- \( L_1: ax + by = 1 \)
- \( L_2: bx + ay = 1 \)

### Step 1: Identify the equations of the lines
The equations of the lines are:
- \( L_1: ax + by - 1 = 0 \)
- \( L_2: bx + ay - 1 = 0 \)

### Step 2: Find the point of intersection (A)
The point A is the intersection of lines \( L_1 \) and \( L_2 \). To find the coordinates of point A, we can solve these two equations simultaneously.

From \( L_1 \):
\[ ax + by = 1 \quad \text{(1)} \]
From \( L_2 \):
\[ bx + ay = 1 \quad \text{(2)} \]

### Step 3: Solve the equations
We can multiply equation (1) by \( b \) and equation (2) by \( a \):
\[ b(ax + by) = b \quad \Rightarrow \quad abx + b^2y = b \quad \text{(3)} \]
\[ a(bx + ay) = a \quad \Rightarrow \quad abx + a^2y = a \quad \text{(4)} \]

Now, subtract equation (4) from equation (3):
\[ (b^2y - a^2y) = b - a \]
\[ (b^2 - a^2)y = b - a \]
\[ y = \frac{b - a}{b^2 - a^2} = \frac{b - a}{(b - a)(b + a)} = \frac{1}{b + a} \quad \text{(if } b \neq a\text{)} \]

Substituting \( y \) back into equation (1) to find \( x \):
\[ ax + b\left(\frac{1}{b + a}\right) = 1 \]
\[ ax + \frac{b}{b + a} = 1 \]
\[ ax = 1 - \frac{b}{b + a} \]
\[ ax = \frac{(b + a) - b}{b + a} = \frac{a}{b + a} \]
\[ x = \frac{1}{b + a} \quad \text{(if } a \neq 0\text{)} \]

Thus, the coordinates of point A are:
\[ A\left(\frac{1}{b + a}, \frac{1}{b + a}\right) \]

### Step 4: Equation of the median AD
The median AD will pass through point A and the midpoint D of BC, which is given as (a, b).

The slope of line AD can be calculated as:
\[ \text{slope} = \frac{b - \frac{1}{b + a}}{a - \frac{1}{b + a}} = \frac{b(b + a) - 1}{a(b + a) - 1} \]

Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = A\left(\frac{1}{b + a}, \frac{1}{b + a}\right) \) and \( m \) is the slope calculated above.

The equation of the median AD becomes:
\[ y - \frac{1}{b + a} = \frac{b(b + a) - 1}{a(b + a) - 1}\left(x - \frac{1}{b + a}\right) \]

### Final Equation
After simplifying, we will arrive at the final equation of the median AD.