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 can follow these steps:

### Step 1: Identify the lines and their intersections
The equations of the lines are:
1. \( y = ax \) (Line 1)
2. \( x + y - a = 0 \) (Line 2)

To find the area of the triangle, we need to determine the points of intersection of these lines with the y-axis and with each other.

### Step 2: Find the intersection of Line 2 with the y-axis
To find the intersection of Line 2 with the y-axis, we set \( x = 0 \):
\[ 0 + y - a = 0 \implies y = a \]
So, the point of intersection with the y-axis is \( B(0, a) \).

### Step 3: Find the intersection of Line 1 and Line 2
To find the intersection of Line 1 and Line 2, we set \( y = ax \) into the equation of Line 2:
\[ x + ax - a = 0 \implies x(1 + a) = a \implies x = \frac{a}{1 + a} \]
Now, substituting \( x \) back into Line 1 to find \( y \):
\[ y = a\left(\frac{a}{1 + a}\right) = \frac{a^2}{1 + a} \]
Thus, the intersection point \( A \) is \( A\left(\frac{a}{1 + a}, \frac{a^2}{1 + a}\right) \).

### Step 4: Identify the vertices of the triangle
The vertices of the triangle formed by the lines and the y-axis are:
- \( O(0, 0) \) (the origin)
- \( A\left(\frac{a}{1 + a}, \frac{a^2}{1 + a}\right) \)
- \( B(0, a) \)

### Step 5: Calculate the area of the triangle
The area \( A \) of a triangle with vertices at \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of the points \( O(0, 0) \), \( A\left(\frac{a}{1 + a}, \frac{a^2}{1 + a}\right) \), and \( B(0, a) \):
\[ \text{Area} = \frac{1}{2} \left| 0\left(\frac{a^2}{1 + a} - a\right) + \frac{a}{1 + a}(a - 0) + 0\left(0 - \frac{a^2}{1 + a}\right) \right| \]
This simplifies to:
\[ = \frac{1}{2} \left| \frac{a^2}{1 + a} \right| = \frac{a^2}{2(1 + a)} \]

### Final Answer
Thus, the area of the triangle formed by the lines is:
\[ \text{Area} = \frac{a^2}{2(1 + a)} \]

---