If area of `DeltaABC` is `12` square units and vertices are `A(x,2)`, `B(4,-1)` and `C(-3,7)` , then find the value of `x`.

Answer

To find the value of \( x \) for the triangle \( \Delta ABC \) with vertices \( A(x, 2) \), \( B(4, -1) \), and \( C(-3, 7) \) and an area of \( 12 \) square units, we can use the formula for the area of a triangle given by its vertices.

### Step 1: Write the formula for the area of the triangle
The area \( A \) of triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our triangle \( A(x, 2) \), \( B(4, -1) \), and \( C(-3, 7) \), we have:
- \( (x_1, y_1) = (x, 2) \)
- \( (x_2, y_2) = (4, -1) \)
- \( (x_3, y_3) = (-3, 7) \)

### Step 2: Substitute the coordinates into the area formula
Substituting the coordinates into the area formula gives:
\[ 12 = \frac{1}{2} \left| x(-1 - 7) + 4(7 - 2) + (-3)(2 + 1) \right| \]
This simplifies to:
\[ 12 = \frac{1}{2} \left| x(-8) + 4(5) - 3(3) \right| \]
\[ 12 = \frac{1}{2} \left| -8x + 20 - 9 \right| \]
\[ 12 = \frac{1}{2} \left| -8x + 11 \right| \]

### Step 3: Eliminate the fraction
Multiplying both sides by \( 2 \) to eliminate the fraction:
\[ 24 = \left| -8x + 11 \right| \]

### Step 4: Set up equations based on the absolute value
This gives us two cases to consider:
1. \( -8x + 11 = 24 \)
2. \( -8x + 11 = -24 \)

### Step 5: Solve the first equation
For the first equation:
\[ -8x + 11 = 24 \]
\[ -8x = 24 - 11 \]
\[ -8x = 13 \]
\[ x = -\frac{13}{8} \]

### Step 6: Solve the second equation
For the second equation:
\[ -8x + 11 = -24 \]
\[ -8x = -24 - 11 \]
\[ -8x = -35 \]
\[ x = \frac{35}{8} \]

### Conclusion
The values of \( x \) that satisfy the area condition are:
\[ x = -\frac{13}{8} \quad \text{and} \quad x = \frac{35}{8} \]