Let's break down the problem step by step.

Understanding the Geometry

• The two triangles share a common vertex at A(0,0).
• Upper Triangle has vertices A(0,0), B(-2,3), and C(2,3).
• Lower Triangle has vertices A(0,0), B'(-1,-3), and C'(1,-3).
• We're asked to find the area of the triangle formed by the intersection of the line BC' and the line B'C.

This involves the geometry of line intersections and computing the area of a triangle formed by the resulting vertices.

Step 1: Find the equations of the lines BC' and B'C

Line BC'

The points B(-2, 3) and C'(1, -3) are given. The slope of the line BC' is calculated using the formula:

$\text{slope of BC'} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 3}{1 - (-2)} = \frac{-6}{3} = -2$

Using point B(-2,3) and the point-slope form of the line equation:

$y - y_1 = m(x - x_1) \implies y - 3 = -2(x + 2)$

Simplifying:

$y - 3 = -2x - 4 \implies y = -2x - 1$

Line B'C

The points B'(-1, -3) and C(2, 3) are given. The slope of the line B'C is:

$\text{slope of B'C} = \frac{3 - (-3)}{2 - (-1)} = \frac{6}{3} = 2$

Using point B'(-1,-3) and the point-slope form of the line equation:

$y - y_1 = m(x - x_1) \implies y + 3 = 2(x + 1)$

Simplifying:

$y + 3 = 2x + 2 \implies y = 2x - 1$

Step 2: Find the Intersection Point A'

The intersection point A' is the solution to the system of equations:

$y = -2x - 1$ $y = 2x - 1$

Setting the two equations equal to each other:

$-2x - 1 = 2x - 1$

Solving for x:

$-2x - 2x = -1 + 1 \implies -4x = 0 \implies x = 0$

Substitute x = 0 into either equation to find y:

$y = -2(0) - 1 = -1$

Thus, the intersection point is A'(0, -1).

Step 3: Area of Triangle A'B'C'

Now, we need to find the area of the triangle formed by A'(0,-1), B'(-1,-3), and C'(1,-3). The formula for the area of a triangle given three vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:

$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Substitute the coordinates A'(0,-1), B'(-1,-3), and C'(1,-3) into the formula:

$\text{Area} = \frac{1}{2} \left| 0((-3) - (-3)) + (-1)((-3) - (-1)) + 1((-1) - (-3)) \right|$ $= \frac{1}{2} \left| 0(0) + (-1)(-2) + 1(2) \right|$ $= \frac{1}{2} \left| 0 + 2 + 2 \right| = \frac{1}{2} \times 4 = 2$

Thus, the area of the triangle formed by A', B', and C' is 2 square units.

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Recap of Steps:

1. Find the equations of lines BC' and B'C.
2. Find the intersection point A'(0, -1) of these two lines.
3. Use the triangle area formula to calculate the area formed by A', B', and C'.

Would you like further clarification or details on any of the steps? Here are some related questions you might find interesting:

1. How do you compute the area of a triangle using determinant form?
2. What happens if the intersection point lies outside the triangle?
3. How can the area change if the coordinates of the vertices are altered?
4. Can we generalize this method for non-linear intersections (curves)?
5. What are some applications of these concepts in real-world geometry problems?

Tip: When finding the intersection of two lines, always check if the slopes are equal first (which means they are parallel) or if they intersect at a unique point!