The vertices of a `DeltaABC` are `(lamda,2-,2lamda),(-lamda+1,2lamda)` and `(-4-lamda,6-2lamda)`. If its area be 70 units, then number of integral values of `lamda` is

Options

A. 1
B. 2
C. 4
D. 0

Answer

To find the number of integral values of \( \lambda \) for the triangle with vertices \( A(\lambda, 2 - 2\lambda) \), \( B(-\lambda + 1, 2\lambda) \), and \( C(-4 - \lambda, 6 - 2\lambda) \) such that its area is 70 units, we can use the formula for the area of a triangle given by its vertices in coordinate geometry.

### Step 1: Area Formula
The area \( A \) of triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our triangle, we have:
- \( A(\lambda, 2 - 2\lambda) \)
- \( B(-\lambda + 1, 2\lambda) \)
- \( C(-4 - \lambda, 6 - 2\lambda) \)

### Step 2: Substitute the Coordinates
Substituting the coordinates into the area formula:
\[ A = \frac{1}{2} \left| \lambda(2\lambda - (6 - 2\lambda)) + (-\lambda + 1)((6 - 2\lambda) - (2 - 2\lambda)) + (-4 - \lambda)((2 - 2\lambda) - 2\lambda) \right| \]

### Step 3: Simplify the Expression
Calculating each term:
1. The first term:
\[ \lambda(2\lambda - 6 + 2\lambda) = \lambda(4\lambda - 6) = 4\lambda^2 - 6\lambda \]

2. The second term:
\[ -\lambda + 1 \text{ gives } (6 - 2\lambda - 2 + 2\lambda) = 4 \implies (-\lambda + 1)(4) = -4\lambda + 4 \]

3. The third term:
\[ -4 - \lambda \text{ gives } (2 - 2\lambda - 2\lambda) = -2\lambda \implies (-4 - \lambda)(-2\lambda) = 8\lambda + 2\lambda^2 \]

Combining all terms:
\[ A = \frac{1}{2} \left| (4\lambda^2 - 6\lambda) + (-4\lambda + 4) + (8\lambda + 2\lambda^2) \right| \]
\[ = \frac{1}{2} \left| 6\lambda^2 + 2\lambda - 2 \right| \]

### Step 4: Set Area to 70
Setting the area equal to 70:
\[ \frac{1}{2} \left| 6\lambda^2 + 2\lambda - 2 \right| = 70 \]
\[ \left| 6\lambda^2 + 2\lambda - 2 \right| = 140 \]

### Step 5: Solve the Absolute Value Equation
This gives us two cases:
1. \( 6\lambda^2 + 2\lambda - 2 = 140 \)
2. \( 6\lambda^2 + 2\lambda - 2 = -140 \)

#### Case 1:
\[ 6\lambda^2 + 2\lambda - 142 = 0 \]
Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 6 \cdot (-142)}}{2 \cdot 6} \]
\[ = \frac{-2 \pm \sqrt{4 + 3408}}{12} = \frac{-2 \pm \sqrt{3412}}{12} \]

#### Case 2:
\[ 6\lambda^2 + 2\lambda + 138 = 0 \]
Using the quadratic formula:
\[ \lambda = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 6 \cdot 138}}{2 \cdot 6} \]
\[ = \frac{-2 \pm \sqrt{4 - 3312}}{12} \text{ (which gives complex roots)} \]

### Step 6: Analyze Integral Values
From Case 1, we need to find the integral values of \( \lambda \) from the roots obtained. The discriminant \( \sqrt{3412} \) is approximately 58.5, leading to two roots:
\[ \lambda_1 = \frac{-2 + 58.5}{12} \approx 4.7 \quad \text{and} \quad \lambda_2 = \frac{-2 - 58.5}{12} \approx -5.0 \]

The integral values of \( \lambda \) between these roots are \( -5, -4, -3, -2, -1, 0, 1, 2, 3, 4 \).

### Conclusion
The total number of integral values of \( \lambda \) is **10**.