The area of the triangle formed by the lines y= ax, x+y-a=0, and y-axis is equal to

Options

A. `(1)/(2|1+a|)`
B. `(a^(2))/(|1+a|)`
C. `(1)/(2)(a)/(|1+a|)`
D. `(a^(2))/(2|1+a|)`

Explanation

To find the area of the triangle formed by the lines \( y = ax \), \( x + y - a = 0 \), and the y-axis, we will follow these steps:

### Step 1: Identify the lines
The equations of the lines are:
1. \( y = ax \) (Equation 1)
2. \( x + y = a \) (Equation 2, rearranged from \( x + y - a = 0 \))

### Step 2: Find the points of intersection
We need to find the points where these lines intersect with each other and with the axes.

#### Intersection of \( y = ax \) and \( x + y = a \):
Substituting \( y = ax \) into \( x + y = a \):
\[ x + ax = a \]
\[ x(1 + a) = a \]
\[ x = \frac{a}{1 + a} \]
Now substituting \( x \) back into Equation 1 to find \( y \):
\[ y = a \left(\frac{a}{1 + a}\right) = \frac{a^2}{1 + a} \]
Thus, the point of intersection \( B \) is:
\[ B\left(\frac{a}{1 + a}, \frac{a^2}{1 + a}\right) \]

### Step 3: Find the points where the lines intersect the axes
- For the line \( y = ax \):
- When \( x = 0 \), \( y = 0 \) (Point \( O(0, 0) \))
- When \( y = 0 \), \( x = 0 \) (Point \( A(a, 0) \))

- For the line \( x + y = a \):
- When \( x = 0 \), \( y = a \) (Point on y-axis)
- When \( y = 0 \), \( x = a \) (Point \( A(a, 0) \))

### Step 4: Calculate the area of triangle \( OAB \)
The area \( A \) of triangle \( OAB \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( OA \) is the distance along the x-axis from \( O(0, 0) \) to \( A(a, 0) \), which is \( a \).

The height is the y-coordinate of point \( B \), which is \( \frac{a^2}{1 + a} \).

Thus, the area becomes:
\[ \text{Area} = \frac{1}{2} \times a \times \frac{a^2}{1 + a} \]
\[ \text{Area} = \frac{a^3}{2(1 + a)} \]

### Final Answer:
The area of the triangle formed by the lines \( y = ax \), \( x + y - a = 0 \), and the y-axis is:
\[ \text{Area} = \frac{a^3}{2(1 + a)} \]