Consider a triangle ABC in the xy -plane with vertices A = (0,0), B = (1,1) and C = (9, 1). If the line x = a divides the triangle into two parts of equal area, then a equals

Options

A. 3
B. 3.5
C. 4
D. 4.5

Answer

To find the value of \( a \) that divides triangle ABC into two parts of equal area, we can follow these steps:

### Step 1: Calculate the Area of Triangle ABC
The vertices of the triangle are given as:
- \( A(0, 0) \)
- \( B(1, 1) \)
- \( C(9, 1) \)

The area \( A \) of a triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

In this triangle:
- The base \( BC \) can be considered as the horizontal distance between points \( B \) and \( C \), which is \( 9 - 1 = 8 \).
- The height from point \( A \) to line \( BC \) is \( 1 \) (the y-coordinate of points \( B \) and \( C \)).

Thus, the area of triangle ABC is:
\[ \text{Area} = \frac{1}{2} \times 8 \times 1 = 4 \text{ square units} \]

### Step 2: Find Half of the Area
Since we want to divide the triangle into two equal areas, we need half of the total area:
\[ \text{Half Area} = \frac{4}{2} = 2 \text{ square units} \]

### Step 3: Set Up the Equation for Area of the New Triangle
We need to find the line \( x = a \) that divides the triangle into two parts of equal area. The area of the triangle formed by points \( A(0, 0) \), \( B(1, 1) \), and the intersection point on line \( x = a \) can be calculated.

The height of the triangle formed will be the vertical distance from line \( BC \) to point \( A \), which is given by:
\[ \text{Height} = 1 - \frac{a}{9} \]
The base of this new triangle will be:
\[ \text{Base} = 9 - a \]

### Step 4: Write the Area Equation
The area of the new triangle can be expressed as:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times (9 - a) \times \left(1 - \frac{a}{9}\right) \]

Setting this equal to \( 2 \) (the half area):
\[ \frac{1}{2} \times (9 - a) \times \left(1 - \frac{a}{9}\right) = 2 \]

### Step 5: Simplify and Solve for \( a \)
Multiply both sides by \( 2 \):
\[ (9 - a) \times \left(1 - \frac{a}{9}\right) = 4 \]

Expanding the left side:
\[ (9 - a) \left(1 - \frac{a}{9}\right) = 9 - a - \frac{9a}{9} + \frac{a^2}{9} = 9 - a - a + \frac{a^2}{9} = 9 - 2a + \frac{a^2}{9} \]

Setting this equal to \( 4 \):
\[ 9 - 2a + \frac{a^2}{9} = 4 \]

Rearranging gives:
\[ \frac{a^2}{9} - 2a + 5 = 0 \]

Multiplying through by \( 9 \) to eliminate the fraction:
\[ a^2 - 18a + 45 = 0 \]

### Step 6: Use the Quadratic Formula
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -18, c = 45 \):
\[ a = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 1 \cdot 45}}{2 \cdot 1} \]
\[ = \frac{18 \pm \sqrt{324 - 180}}{2} \]
\[ = \frac{18 \pm \sqrt{144}}{2} \]
\[ = \frac{18 \pm 12}{2} \]

Calculating the two possible values:
1. \( a = \frac{30}{2} = 15 \)
2. \( a = \frac{6}{2} = 3 \)

Since \( a \) must be between \( 0 \) and \( 9 \) (the x-coordinates of points \( A \) and \( C \)), we take:
\[ a = 3 \]

### Final Answer
Thus, the value of \( a \) that divides the triangle into two equal areas is:
\[ \boxed{3} \]