Drawn from origin there are two mutually perpendicular lines forming an isosceles triangle together with the straight line `2x+y=5` then the area of this triangle is

Answer

To solve the problem of finding the area of the isosceles triangle formed by two mutually perpendicular lines from the origin and the line \(2x + y = 5\), we can follow these steps:

### Step 1: Identify the lines
The two lines drawn from the origin are mutually perpendicular. We can represent them as:
- Line 1: \(y = mx\) (where \(m\) is the slope)
- Line 2: \(y = -\frac{1}{m}x\) (the slope is the negative reciprocal of \(m\))

### Step 2: Find the intersection points
The intersection of the line \(2x + y = 5\) with the two lines from the origin will give us the vertices of the triangle.

### Step 3: Calculate the coordinates of the intersection points
To find the intersection points, we substitute \(y\) from the line equations into \(2x + y = 5\).

1. For Line 1: \(y = mx\)
\[ 2x + mx = 5 \implies x(2 + m) = 5 \implies x = \frac{5}{2 + m} \]
\[ y = m \left(\frac{5}{2 + m}\right) = \frac{5m}{2 + m} \]
Thus, the intersection point \(A\) is \(\left(\frac{5}{2 + m}, \frac{5m}{2 + m}\right)\).

2. For Line 2: \(y = -\frac{1}{m}x\)
\[ 2x - \frac{1}{m}x = 5 \implies x(2 - \frac{1}{m}) = 5 \implies x = \frac{5m}{2m - 1} \]
\[ y = -\frac{1}{m}\left(\frac{5m}{2m - 1}\right) = -\frac{5}{2m - 1} \]
Thus, the intersection point \(B\) is \(\left(\frac{5m}{2m - 1}, -\frac{5}{2m - 1}\right)\).

### Step 4: Calculate the height of the triangle
The height of the triangle from the origin \(O(0, 0)\) to the line \(2x + y = 5\) can be calculated using the formula for the distance from a point to a line:
\[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(2x + y - 5 = 0\) (where \(A = 2\), \(B = 1\), and \(C = -5\)):
\[ \text{Distance from } O(0, 0) = \frac{|2(0) + 1(0) - 5|}{\sqrt{2^2 + 1^2}} = \frac{5}{\sqrt{5}} = \sqrt{5} \]

### Step 5: Calculate the base of the triangle
The base \(AB\) can be calculated as the distance between points \(A\) and \(B\). However, since both points are symmetrical with respect to the origin and the line, we can say that:
\[ AB = OA + OB = 2 \cdot OA \]
Where \(OA\) is the distance from the origin to point \(A\).

### Step 6: Calculate the area of the triangle
The area \(A\) of triangle \(OAB\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Substituting the values:
\[ \text{Area} = \frac{1}{2} \times 2 \cdot OA \times \sqrt{5} = OA \cdot \sqrt{5} \]

### Step 7: Final Calculation
Since \(OA = \sqrt{5}\):
\[ \text{Area} = \sqrt{5} \cdot \sqrt{5} = 5 \]

Thus, the area of the triangle formed is \(5\).