Let `triangleABC` is a right angled triangle right angled at A such that `A(1,2),C(3,1)` and area of `triangleABC=5sqrt(5)` then abscissa of B can be

Options

A. `1+5sqrt(2)`
B. `1+2sqrt(5)`
C. `1-5sqrt(2)`
D. `3+2sqrt(5)`

Answer

To solve the problem, we need to find the abscissa (x-coordinate) of point B in triangle ABC, which is right-angled at A. The coordinates of points A and C are given as A(1, 2) and C(3, 1), and the area of triangle ABC is given as \(5\sqrt{5}\).

### Step 1: Set up the coordinates
Let the coordinates of point B be \(B(\alpha, \beta)\). We know:
- A = (1, 2)
- C = (3, 1)

### Step 2: Calculate the slope of AC
The slope of line AC can be calculated using the formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]
For points A(1, 2) and C(3, 1):
\[ \text{slope of AC} = \frac{1 - 2}{3 - 1} = \frac{-1}{2} \]

### Step 3: Use the property of perpendicular lines
Since triangle ABC is right-angled at A, the slope of line AB (M_AB) and the slope of line AC (M_AC) must satisfy:
\[ M_{AB} \cdot M_{AC} = -1 \]
Let \(M_{AB} = \frac{\beta - 2}{\alpha - 1}\). Thus:
\[ \frac{\beta - 2}{\alpha - 1} \cdot \left(-\frac{1}{2}\right) = -1 \]
This simplifies to:
\[ \frac{\beta - 2}{\alpha - 1} = 2 \]
From this, we can express \(\beta\) in terms of \(\alpha\):
\[ \beta - 2 = 2(\alpha - 1) \implies \beta = 2\alpha - 2 + 2 \implies \beta = 2\alpha \]

### Step 4: Calculate the area of triangle ABC
The area of triangle ABC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can use the lengths of sides AB and AC. The area is given as \(5\sqrt{5}\):
\[ \frac{1}{2} \times AB \times AC = 5\sqrt{5} \]
Thus:
\[ AB \times AC = 10\sqrt{5} \]

### Step 5: Calculate lengths AB and AC
Using the distance formula:
\[ AB = \sqrt{(\alpha - 1)^2 + (\beta - 2)^2} \]
Substituting \(\beta = 2\alpha\):
\[ AB = \sqrt{(\alpha - 1)^2 + (2\alpha - 2)^2} \]
Calculating \(AC\):
\[ AC = \sqrt{(3 - 1)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 6: Substitute into area equation
Now substituting into the area equation:
\[ \sqrt{(\alpha - 1)^2 + (2\alpha - 2)^2} \cdot \sqrt{5} = 10\sqrt{5} \]
Dividing both sides by \(\sqrt{5}\):
\[ \sqrt{(\alpha - 1)^2 + (2\alpha - 2)^2} = 10 \]

### Step 7: Square both sides
Squaring both sides gives:
\[ (\alpha - 1)^2 + (2\alpha - 2)^2 = 100 \]
Expanding:
\[ (\alpha - 1)^2 + 4(\alpha - 1)^2 = 100 \]
\[ 5(\alpha - 1)^2 = 100 \implies (\alpha - 1)^2 = 20 \]
Taking the square root:
\[ \alpha - 1 = \pm \sqrt{20} \implies \alpha = 1 \pm 2\sqrt{5} \]

### Step 8: Final values for \(\alpha\)
Thus, the possible values for the abscissa of point B are:
\[ \alpha = 1 + 2\sqrt{5} \quad \text{or} \quad \alpha = 1 - 2\sqrt{5} \]