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

Options

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

Answer

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

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

### Step 2: Find the coordinates of point A
The line AB is given by the equation \( y = \frac{x}{2} \). We can express the coordinates of point A as \( (a, \frac{a}{2}) \), where \( a \) is the x-coordinate of point A.

### Step 3: Use the distance formula to set up the equation
Since triangle ABC is isosceles, the lengths AB and AC must be equal. We can use the distance formula to express these lengths:

- Length of AB:
\[ AB = \sqrt{(a - 2)^2 + \left(\frac{a}{2} - 1\right)^2} \]

- Length of AC:
\[ AC = \sqrt{(a - 1)^2 + \left(\frac{a}{2} - 2\right)^2} \]

### Step 4: Set the lengths equal to each other
Since \( AB = AC \), we can set the two expressions equal:
\[ \sqrt{(a - 2)^2 + \left(\frac{a}{2} - 1\right)^2} = \sqrt{(a - 1)^2 + \left(\frac{a}{2} - 2\right)^2} \]

### Step 5: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ (a - 2)^2 + \left(\frac{a}{2} - 1\right)^2 = (a - 1)^2 + \left(\frac{a}{2} - 2\right)^2 \]

### Step 6: Expand both sides
Expanding both sides:
- Left Side:
\[ (a - 2)^2 = a^2 - 4a + 4 \]
\[ \left(\frac{a}{2} - 1\right)^2 = \frac{a^2}{4} - a + 1 \]
Combining these:
\[ a^2 - 4a + 4 + \frac{a^2}{4} - a + 1 = \frac{5a^2}{4} - 5a + 5 \]

- Right Side:
\[ (a - 1)^2 = a^2 - 2a + 1 \]
\[ \left(\frac{a}{2} - 2\right)^2 = \frac{a^2}{4} - 2a + 4 \]
Combining these:
\[ a^2 - 2a + 1 + \frac{a^2}{4} - 2a + 4 = \frac{5a^2}{4} - 4a + 5 \]

### Step 7: Set the expanded sides equal
Now we have:
\[ \frac{5a^2}{4} - 5a + 5 = \frac{5a^2}{4} - 4a + 5 \]

### Step 8: Simplify the equation
Subtract \(\frac{5a^2}{4} + 5\) from both sides:
\[ -5a = -4a \]
This simplifies to:
\[ -a = 0 \implies a = 0 \]

### Step 9: Find the coordinates of point A
Substituting \( a = 0 \) back into the coordinates of A:
\[ A = (0, \frac{0}{2}) = (0, 0) \]

### Step 10: Find the equation of line AC
Now, we can find the equation of line AC using points A (0, 0) and C (1, 2):
Using the slope formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{1 - 0} = 2 \]
Using point-slope form:
\[ y - 0 = 2(x - 0) \implies y = 2x \]

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