The line x = c cuts the triangle with corners (0,0) , (1,1) and (9,1) into two regions .For the area of the two regions to the same , then c must be equal to

Options

A. `5/2`
B. 3
C. `7/2`
D. 3 or 15

Answer

To find the value of \( c \) such that the line \( x = c \) divides the triangle with vertices at \( (0,0) \), \( (1,1) \), and \( (9,1) \) into two regions of equal area, we can follow these steps:

### Step 1: Calculate the Area of the Triangle
The area \( A \) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In our triangle, the base can be taken as the distance between points \( (1,1) \) and \( (9,1) \), which is \( 9 - 1 = 8 \). The height is the vertical distance from the base (y = 1) to the vertex at \( (0,0) \), which is \( 1 \).

Thus, the area of triangle \( ABC \) is:
\[ A = \frac{1}{2} \times 8 \times 1 = 4 \]

### Step 2: Set Up the Area Condition
Since we want the line \( x = c \) to divide the triangle into two equal areas, each region must have an area of:
\[ \frac{4}{2} = 2 \]

### Step 3: Determine the Area of Triangle \( CDE \)
The triangle formed by the line \( x = c \) and the points \( (c, 0) \) and \( (c, 1) \) will have its area calculated as follows. The base of triangle \( CDE \) is \( 9 - c \) (the horizontal distance from \( (c, 1) \) to \( (9, 1) \)), and the height is \( 1 - \frac{c}{9} \) (the vertical distance from the line \( y = 1 \) to the line connecting \( (0,0) \) and \( (9,1) \)).

Thus, the area of triangle \( CDE \) is:
\[ \text{Area}_{CDE} = \frac{1}{2} \times (9 - c) \times \left(1 - \frac{c}{9}\right) \]

### Step 4: Set the Area Equal to 2
We set the area of triangle \( CDE \) equal to 2:
\[ \frac{1}{2} \times (9 - c) \times \left(1 - \frac{c}{9}\right) = 2 \]
Multiplying both sides by 2 gives:
\[ (9 - c) \times \left(1 - \frac{c}{9}\right) = 4 \]

### Step 5: Expand and Rearrange the Equation
Expanding the left side:
\[ (9 - c) \left(1 - \frac{c}{9}\right) = 9 - c - \frac{9c}{9} + \frac{c^2}{9} = 9 - c - c + \frac{c^2}{9} = 9 - 2c + \frac{c^2}{9} \]
Setting this equal to 4:
\[ 9 - 2c + \frac{c^2}{9} = 4 \]
Rearranging gives:
\[ \frac{c^2}{9} - 2c + 5 = 0 \]

### Step 6: Multiply Through by 9
To eliminate the fraction, multiply the entire equation by 9:
\[ c^2 - 18c + 45 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ c = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 1 \cdot 45}}{2 \cdot 1} \]
Calculating the discriminant:
\[ c = \frac{18 \pm \sqrt{324 - 180}}{2} = \frac{18 \pm \sqrt{144}}{2} = \frac{18 \pm 12}{2} \]
This gives us two solutions:
\[ c = \frac{30}{2} = 15 \quad \text{and} \quad c = \frac{6}{2} = 3 \]

### Final Answer
Thus, the values of \( c \) that divide the triangle into two equal areas are:
\[ c = 3 \quad \text{and} \quad c = 15 \]