Let `Delta ABC` is an isosceles trianlge with `AB = AC`. If `B = (0,a) , C = (2a, 0)` and the equation of AB is `3x - 4y + 4a = 0`, then the equation of side AC is

Options

A. `y = 8x - 16 a`
B. `3y = 4x - 8a`
C. `x = 2a`
D. `y + 8x = 16 a`

Answer

To find the equation of side AC in the isosceles triangle ABC, we will follow these steps:

### Step 1: Identify the coordinates of points B and C
Given:
- \( B = (0, a) \)
- \( C = (2a, 0) \)

### Step 2: Find the slope of line AB
The equation of line AB is given as:
\[ 3x - 4y + 4a = 0 \]

We can rearrange this equation into the slope-intercept form \( y = mx + b \):
\[ 4y = 3x + 4a \implies y = \frac{3}{4}x + a \]
Thus, the slope \( m_{AB} \) of line AB is \( \frac{3}{4} \).

### Step 3: Find the slope of line BC
To find the slope of line BC, we can use the coordinates of points B and C:
\[ \text{slope of } BC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - a}{2a - 0} = \frac{-a}{2a} = -\frac{1}{2} \]

### Step 4: Use the property of isosceles triangles
In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, the angle between line AB and line BC is equal to the angle between line AC and line BC.

Using the tangent of the angle between two lines:
\[ \tan(\theta_{AB, BC}) = \left| \frac{m_{AB} - m_{BC}}{1 + m_{AB} \cdot m_{BC}} \right| \]
Substituting the slopes:
\[ \tan(\theta_{AB, BC}) = \left| \frac{\frac{3}{4} - \left(-\frac{1}{2}\right)}{1 + \frac{3}{4} \cdot \left(-\frac{1}{2}\right)} \right| \]
Calculating the numerator:
\[ \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} \]
Calculating the denominator:
\[ 1 - \frac{3}{8} = \frac{8}{8} - \frac{3}{8} = \frac{5}{8} \]
Thus,
\[ \tan(\theta_{AB, BC}) = \left| \frac{\frac{5}{4}}{\frac{5}{8}} \right| = \left| \frac{5}{4} \cdot \frac{8}{5} \right| = 2 \]

### Step 5: Find the slope of line AC
Let the slope of line AC be \( m_{AC} \). Since the angles are equal, we have:
\[ \tan(\theta_{AC, BC}) = \tan(\theta_{AB, BC}) = 2 \]

### Step 6: Use the point-slope form to find the equation of line AC
Using point C (which is \( (2a, 0) \)) and the slope \( m_{AC} = 2 \):
\[ y - 0 = 2(x - 2a) \]
Simplifying this gives:
\[ y = 2x - 4a \]

### Step 7: Write the equation in standard form
Rearranging gives us:
\[ 2x - y - 4a = 0 \]

### Final Answer
The equation of side AC is:
\[ 2x - y - 4a = 0 \]