The vertices of a triangle ABC are `(lamda, 2-2lamda), (-lamda+1,2lamda)` an `(_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 given triangle vertices, we will follow these steps:

### Step 1: Write down the vertices of the triangle
The vertices of triangle \(ABC\) are given as:
- \(A(\lambda, 2 - 2\lambda)\)
- \(B(-\lambda + 1, 2\lambda)\)
- \(C(-4 - \lambda, 6 - 2\lambda)\)

### Step 2: Use the formula for the area of a triangle
The area \(A\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of points \(A\), \(B\), and \(C\):
\[ A = \frac{1}{2} \left| \lambda(2\lambda - (6 - 2\lambda)) + (-\lambda + 1)((6 - 2\lambda) - (2 - 2\lambda)) + (-4 - \lambda)((2 - 2\lambda) - 2\lambda) \right| \]

### Step 3: Simplify the expression
Calculating each term:
1. \(A = \frac{1}{2} \left| \lambda(2\lambda - 6 + 2\lambda) \right| = \frac{1}{2} \left| \lambda(4\lambda - 6) \right|\)
2. \(B = -\lambda + 1 \Rightarrow (6 - 2\lambda - 2 + 2\lambda) = 4\)
3. \(C = -4 - \lambda \Rightarrow (2 - 2\lambda - 2\lambda) = -2\lambda\)

Putting it all together:
\[ A = \frac{1}{2} \left| \lambda(4\lambda - 6) + (-\lambda + 1)(4) + (-4 - \lambda)(-2\lambda) \right| \]

### Step 4: Set the area equal to 70
We know the area is given as 70 units:
\[ \frac{1}{2} \left| \lambda(4\lambda - 6) + 4(-\lambda + 1) + 2\lambda(4 + \lambda) \right| = 70 \]
Multiplying both sides by 2:
\[ \left| \lambda(4\lambda - 6) + 4(-\lambda + 1) + 2\lambda(4 + \lambda) \right| = 140 \]

### Step 5: Solve the equation
This leads to two cases:
1. \( \lambda(4\lambda - 6) + 4(-\lambda + 1) + 2\lambda(4 + \lambda) = 140 \)
2. \( \lambda(4\lambda - 6) + 4(-\lambda + 1) + 2\lambda(4 + \lambda) = -140 \)

### Step 6: Simplify and solve for \(\lambda\)
For the first case:
\[ 2\lambda^2 + \lambda - 36 = 0 \]
Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 288}}{4} = \frac{-1 \pm 17}{4} \]
This gives:
\[ \lambda = 4 \quad \text{or} \quad \lambda = -\frac{9}{2} \]

For the second case:
\[ 2\lambda^2 + \lambda + 36 = 0 \]
The discriminant is negative, hence no real solutions.

### Step 7: Count integral values
The integral values of \(\lambda\) from the first case are:
- \(\lambda = 4\) (integral)
- \(\lambda = -4\) (integral)

Thus, the number of integral values of \(\lambda\) is **1**.

### Final Answer
The number of integral values of \(\lambda\) is **1**.

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