Let the area of the triangle with vertices `A(1,alpha) ,B(alpha,0)and C(0,alpha)` be 4 sq. units If the points `(alpha,-alpha),(-alpha,alpha)and (alpha^(2),beta)` are collinear , then `beta ` is equal to

Options

A. 64
B. `-8`
C. `-64`
D. 512

Answer

To solve the problem step by step, we will first find the value of \(\alpha\) using the area of the triangle formed by the points \(A(1, \alpha)\), \(B(\alpha, 0)\), and \(C(0, \alpha)\). Then, we will use the collinearity condition to find the value of \(\beta\).

### Step 1: Calculate the area of the triangle
The area \(A\) of a triangle formed by three points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:

\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

For our triangle \(A(1, \alpha)\), \(B(\alpha, 0)\), and \(C(0, \alpha)\):

- \(x_1 = 1\), \(y_1 = \alpha\)
- \(x_2 = \alpha\), \(y_2 = 0\)
- \(x_3 = 0\), \(y_3 = \alpha\)

Substituting these values into the area formula gives:

\[ A = \frac{1}{2} \left| 1(0 - \alpha) + \alpha(\alpha - \alpha) + 0(\alpha - 0) \right| \]
\[ = \frac{1}{2} \left| -\alpha \right| = \frac{\alpha}{2} \]

According to the problem, the area is equal to 4 square units:

\[ \frac{\alpha}{2} = 4 \]

### Step 2: Solve for \(\alpha\)
Multiplying both sides by 2:

\[ \alpha = 8 \]

### Step 3: Check the collinearity condition
The points \((\alpha, -\alpha)\), \((- \alpha, \alpha)\), and \((\alpha^2, \beta)\) are collinear. The area of the triangle formed by these points must be zero.

Using the same area formula, we have:

- \(x_1 = \alpha\), \(y_1 = -\alpha\)
- \(x_2 = -\alpha\), \(y_2 = \alpha\)
- \(x_3 = \alpha^2\), \(y_3 = \beta\)

Substituting these values into the area formula gives:

\[ 0 = \frac{1}{2} \left| \alpha(\alpha - \beta) + (-\alpha)(\beta + \alpha) + \alpha^2(-\alpha - \alpha) \right| \]
\[ = \frac{1}{2} \left| \alpha^2 - \alpha \beta - \alpha \beta - \alpha^2 - 2\alpha^3 \right| \]
\[ = \frac{1}{2} \left| -2\alpha \beta - 2\alpha^3 \right| \]

Setting this equal to zero gives:

\[ -2\alpha \beta - 2\alpha^3 = 0 \]

### Step 4: Solve for \(\beta\)
Factoring out \(-2\):

\[ \alpha \beta + \alpha^3 = 0 \]
\[ \beta = -\alpha^2 \]

Substituting \(\alpha = 8\):

\[ \beta = -8^2 = -64 \]

### Final Answer
Thus, the value of \(\beta\) is:

\[ \beta = -64 \]