Draw the graph of the linear equation 2x + 3y = 12. Find the coordinates of the
points where the graph of this equation intersects x-axis and y-axis. Shade the area
bounded by the graph of the equation and both the axes.
Find the area of the triangle formed by the graph of the above equation cuts the x-axis
and the y-axis.

### 4. Area of the Triangle Formed

The area of the triangle formed by the graph of the equation \(2x + 3y = 12\) with the x-axis and y-axis can be calculated using the formula for the area of a triangle:

\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]

- The base of the triangle is the distance along the x-axis, which is 6 units (from \((6, 0)\) to \((0, 0)\)).
- The height of the triangle is the distance along the y-axis, which is 4 units (from \((0, 4)\) to \((0, 0)\)).

Thus, the area is:

\[
\text{Area} = \frac{1}{2} \times 6 \times 4 = 12 \text{ square units}
\]

Would you like more details or have any questions? Here are some related questions to consider:

1. How would the graph change if the equation were \(2x + 3y = 6\) instead?
2. What happens to the area of the triangle if the coefficients of x and y in the equation change?
3. How would you find the area if the equation was quadratic instead of linear?
4. Can you find the coordinates of another point on the line, not on the axes?
5. How can you verify that the area calculated geometrically matches algebraic methods?

**Tip:** Understanding the impact of coefficients in linear equations on the graph's slope can help predict changes in intersections with the axes.

Explore the graph of the linear equation 2x + 3y = 12 and calculate the area of the triangle formed by its intersections with the x-axis and y-axis. This problem covers key concepts in linear equations, coordinate geometry, and triangle area calculations. Suitable for Grades 7-9.

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