Find the perimeter of the triangle formed at the intersection of the straight line 3x + 4y = 24 with the coordinate axes

Find the coordinates of the intersection points of the straight line given by the equation

3 * x + 4 * y = 24,

with coordinate axes.

This line intersects the Y-axis at x = 0. Therefore,

3 * 0 + 4 * y = 24,

4 * y = 24,

y = 6.

Intersection point of the straight line with the Y-axis: A (0, 6).

This line intersects the X-axis at y = 0. Therefore,

3 * x + 4 * 0 = 24,

3 * x = 24,

x = 8.

Intersection point of the straight line with the Y-axis: B (8, 0).

The triangle, which is formed by a given line and coordinate axes, has vertices

with coordinates:

O (0, 0), A (0, 6), B (8, 0).

Distance between vertices A and B:

AB = √6 ^ 2 + 8 ^ 2 = √36 + 64 = 10.

Therefore, the perimeter P of triangle ABO is:

P = OA + AB + OB = 6 + 10 + 8 = 24.

Answer: P = 24.