A triangle has two of its vertices at `(0,1)` and `(2,2)` in the cartesian plane. Its third vertex lies on the x-axis. If the area of the triangle is 2 square units then the sum of the possible abscissae of the third vertex, is-

Options

A. 40
B. 10
C. 52
D. 61

Answer

To solve the problem step by step, we will follow the given information and apply the formula for the area of a triangle.

### Step 1: Identify the vertices of the triangle
We have two vertices of the triangle:
- Vertex A: \( (0, 1) \)
- Vertex B: \( (2, 2) \)

The third vertex, which lies on the x-axis, can be represented as:
- Vertex C: \( (h, 0) \)

### Step 2: Use the formula for the area of a triangle
The area \( A \) of a triangle formed by three vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 3: Substitute the coordinates into the area formula
Substituting the coordinates of the vertices into the area formula:
- \( x_1 = 0, y_1 = 1 \)
- \( x_2 = 2, y_2 = 2 \)
- \( x_3 = h, y_3 = 0 \)

The area becomes:
\[ A = \frac{1}{2} \left| 0(2 - 0) + 2(0 - 1) + h(1 - 2) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| 0 + 2(-1) + h(-1) \right| = \frac{1}{2} \left| -2 - h \right| \]

### Step 4: Set the area equal to 2 square units
According to the problem, the area of the triangle is 2 square units. Therefore, we set up the equation:
\[ \frac{1}{2} \left| -2 - h \right| = 2 \]

### Step 5: Solve for \( h \)
Multiplying both sides by 2 gives:
\[ \left| -2 - h \right| = 4 \]

This absolute value equation can be split into two cases:
1. \( -2 - h = 4 \)
2. \( -2 - h = -4 \)

#### Case 1:
\[ -2 - h = 4 \implies -h = 6 \implies h = -6 \]

#### Case 2:
\[ -2 - h = -4 \implies -h = -2 \implies h = 2 \]

### Step 6: Find the sum of the possible abscissae
The possible values for \( h \) are \( -6 \) and \( 2 \). The sum of the possible abscissae is:
\[ -6 + 2 = -4 \]

### Final Answer
The sum of the possible abscissae of the third vertex is \( -4 \).

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