The area of the triangle formed between lines x = 0, y = 0 and 2x – 3y + 6 = 0 is

To find the area of the triangle formed by the lines x = 0, y = 0 and 2x – 3y + 6 = 0, we first need to find the points of intersection of these lines.

1. The line x = 0 intersects the line 2x – 3y + 6 = 0 when x = 0. Substituting x = 0 into the equation 2x – 3y + 6 = 0 gives us -3y + 6 = 0, which simplifies to y = 2. So, the point of intersection is (0, 2).

2. The line y = 0 intersects the line 2x – 3y + 6 = 0 when y = 0. Substituting y = 0 into the equation 2x – 3y + 6 = 0 gives us 2x + 6 = 0, which simplifies to x = -3. So, the point of intersection is (-3, 0).

3. The lines x = 0 and y = 0 intersect at the origin (0, 0).

So, the vertices of the triangle are (0, 2), (-3, 0), and (0, 0).

The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by the formula:

Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Substituting the coordinates of the vertices into this formula gives:

Area = 1/2 * |0*(0 - 0) + (-3)(2 - 0) + 0(0 - 2)| = 1/2 * |-6| = 3 square units.