In an isosceles triangle `ABC,` the coordinates of the points `B`and `C` on the base `BC` are respectively `(1, 2)` and `(2. 1).` If the equation of the line `AB` is `y= 2x,` then the equation of the line `AC` is

Options

A. `y = (1)/(2) (x-1)`
B. `y = (x)/(2)`
C. `y = x - 1`
D. `2y = x +3`

Explanation

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

### Step 1: Identify the coordinates of points B and C
The coordinates of points B and C are given as:
- \( B(1, 2) \)
- \( C(2, 1) \)

### Step 2: Determine the coordinates of point A
Since triangle ABC is isosceles, the distances from point A to points B and C must be equal. We can express point A in terms of its coordinates. Let's denote the coordinates of point A as \( A(a, 2a) \).

### Step 3: Set up the distance equations
Using the distance formula, the distance from A to B must equal the distance from A to C:
\[ AB = AC \]
Calculating \( AB \):
\[ AB = \sqrt{(a - 1)^2 + (2a - 2)^2} \]
Calculating \( AC \):
\[ AC = \sqrt{(a - 2)^2 + (2a - 1)^2 \]
Setting these distances equal gives us:
\[ \sqrt{(a - 1)^2 + (2a - 2)^2} = \sqrt{(a - 2)^2 + (2a - 1)^2} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides results in:
\[ (a - 1)^2 + (2a - 2)^2 = (a - 2)^2 + (2a - 1)^2 \]

### Step 5: Expand both sides
Expanding both sides:
\[ (a^2 - 2a + 1) + (4a^2 - 8a + 4) = (a^2 - 4a + 4) + (4a^2 - 4a + 1) \]
This simplifies to:
\[ 5a^2 - 10a + 5 = 5a^2 - 8a + 5 \]

### Step 6: Simplify the equation
Subtract \( 5a^2 + 5 \) from both sides:
\[ -10a = -8a \]
This leads to:
\[ -2a = 0 \implies a = 0 \]
Thus, the coordinates of point A are:
\[ A(0, 0) \]

### Step 7: Find the slope of line AC
Now we need to find the slope of line AC. The coordinates of points A and C are:
- \( A(0, 0) \)
- \( C(2, 1) \)

The slope \( m \) of line AC is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 0}{2 - 0} = \frac{1}{2} \]

### Step 8: Write the equation of line AC
Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \):
\[ y - 1 = \frac{1}{2}(x - 2) \]
Expanding this gives:
\[ 2y - 2 = x - 2 \]
Rearranging to the standard form:
\[ 2y = x \]
Thus, the equation of line AC is:
\[ y = \frac{1}{2}x \]

### Final Answer
The equation of line AC is:
\[ y = \frac{1}{2}x \]