Drawn from origin are two mutually perpendicular lines forming an isosceles triangle together with the straight line `2x+y=a` then the area of this triangle is

Options

A. `(a^(2))/2` sq units
B. `(a^(2))/3` sq units
C. `(a^(2))/5` sq units
D. None of these

Answer

To solve the problem of finding the area of the isosceles triangle formed by two mutually perpendicular lines drawn from the origin and the line \(2x + y = a\), we can follow these steps:

### Step 1: Understand the Geometry
The two mutually perpendicular lines from the origin can be represented as the x-axis and y-axis. This means the triangle is formed by the points (0, 0), (a/2, 0), and (0, a) where the line intersects the axes.

### Step 2: Find the Intercepts of the Line
To find the area of the triangle, we first need to determine where the line \(2x + y = a\) intersects the x-axis and y-axis.

- **X-intercept**: Set \(y = 0\):
\[ 2x + 0 = a \implies x = \frac{a}{2} \]
So the x-intercept is \(\left(\frac{a}{2}, 0\right)\).

- **Y-intercept**: Set \(x = 0\):
\[ 2(0) + y = a \implies y = a \]
So the y-intercept is \((0, a)\).

### Step 3: Identify the Vertices of the Triangle
The vertices of the triangle formed by the origin and the intercepts are:
- Vertex A: (0, 0) (the origin)
- Vertex B: \(\left(\frac{a}{2}, 0\right)\) (x-intercept)
- Vertex C: (0, a) (y-intercept)

### Step 4: Calculate the Area of the Triangle
The area \(A\) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In our case, the base is the x-intercept \(\frac{a}{2}\) and the height is the y-intercept \(a\).

Substituting these values into the area formula:
\[ A = \frac{1}{2} \times \left(\frac{a}{2}\right) \times a = \frac{1}{2} \times \frac{a^2}{2} = \frac{a^2}{4} \]

### Final Answer
The area of the triangle is:
\[ \boxed{\frac{a^2}{4}} \]