The equation of AB,BC, AC three sides of a triangle are `-x + y -1 =0, x + y -1 =0 and x =-4if (alpha , 0) and (0, beta)` lie inside the triangle where `alpha, beta in Z,` then

Options

A. `alpha =-3,-2 beta in phi`
B. `alpha =-3, -2, beta =1`
C. Orthocentre of triangle ABC is at `(0,1)`
D. The circumentre of the triangle will lie on the line `x =1 `

Answer

To solve the problem, we need to analyze the equations of the sides of the triangle and find the coordinates of the vertices formed by these lines. Then, we will determine the integer values of \( \alpha \) and \( \beta \) that lie inside the triangle.

### Step 1: Identify the equations of the sides of the triangle
The equations of the sides of the triangle are:
1. \( -x + y - 1 = 0 \) (or \( y = x + 1 \))
2. \( x + y - 1 = 0 \) (or \( y = -x + 1 \))
3. \( x = -4 \)

### Step 2: Find the points of intersection (vertices of the triangle)
To find the vertices of the triangle, we will calculate the intersection points of the lines.

**Intersection of Line 1 and Line 2:**
\[ -x + y - 1 = 0 \quad \text{(1)} \]
\[ x + y - 1 = 0 \quad \text{(2)} \]
From (1), we can express \( y \) in terms of \( x \):
\[ y = x + 1 \]
Substituting \( y \) in (2):
\[ x + (x + 1) - 1 = 0 \implies 2x = 0 \implies x = 0 \]
Substituting \( x = 0 \) back into \( y = x + 1 \):
\[ y = 0 + 1 = 1 \]
Thus, the first vertex \( A(0, 1) \).

**Intersection of Line 1 and Line 3:**
Substituting \( x = -4 \) into \( y = x + 1 \):
\[ y = -4 + 1 = -3 \]
Thus, the second vertex \( B(-4, -3) \).

**Intersection of Line 2 and Line 3:**
Substituting \( x = -4 \) into \( y = -x + 1 \):
\[ y = -(-4) + 1 = 4 + 1 = 5 \]
Thus, the third vertex \( C(-4, 5) \).

### Step 3: Determine the area of the triangle
The vertices of the triangle are \( A(0, 1) \), \( B(-4, -3) \), and \( C(-4, 5) \).

To find the area of triangle \( ABC \) 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:
\[ \text{Area} = \frac{1}{2} \left| 0(-3 - 5) + (-4)(5 - 1) + (-4)(1 - (-3)) \right| \]
\[ = \frac{1}{2} \left| 0 + (-4)(4) + (-4)(4) \right| \]
\[ = \frac{1}{2} \left| -16 - 16 \right| = \frac{1}{2} \times 32 = 16 \]

### Step 4: Determine the integer values of \( \alpha \) and \( \beta \)
The points \( (\alpha, 0) \) and \( (0, \beta) \) must lie inside the triangle.

**For \( \alpha \):**
The x-coordinates of the triangle are from \( -4 \) to \( 0 \). Thus, \( \alpha \) must satisfy:
\[ -4 < \alpha < 0 \]
The integer values for \( \alpha \) are \( -3, -2, -1 \).

**For \( \beta \):**
The y-coordinates of the triangle are from \( -3 \) to \( 5 \). Thus, \( \beta \) must satisfy:
\[ -3 < \beta < 5 \]
The integer values for \( \beta \) are \( -2, -1, 0, 1, 2, 3, 4 \).

### Final Answer
The integer values of \( \alpha \) are \( -3, -2, -1 \) and for \( \beta \) are \( -2, -1, 0, 1, 2, 3, 4 \).