The plane x/1+y/2+z/−1=1 meets the coordinate axes in the points A,B and C respectively then area of triangle ABC is

To find the area of triangle ABC, we first need to determine the coordinates of points A, B, and C where the plane intersects the coordinate axes.

Let's start with point A. To find the x-coordinate of point A, we set y and z to 0 in the equation of the plane:

x/1 + 0/2 + 0/(-1) = 1 x/1 = 1 x = 1

So, point A has coordinates (1, 0, 0).

Next, let's find the coordinates of point B. We set x and z to 0 in the equation of the plane:

0/1 + y/2 + 0/(-1) = 1 y/2 = 1 y = 2

Therefore, point B has coordinates (0, 2, 0).

Finally, we find the coordinates of point C by setting x and y to 0 in the equation of the plane:

0/1 + 0/2 + z/(-1) = 1 z/(-1) = 1 z = -1

Hence, point C has coordinates (0, 0, -1).

Now that we have the coordinates of points A, B, and C, we can calculate the area of triangle ABC using the formula for the area of a triangle given its vertices.

The formula for the area of a triangle with vertices (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) is:

Area = 1/2 * |x1(y2z3 - y3z2) + x2(y3z1 - y1z3) + x3(y1z2 - y2z1)|

Plugging in the coordinates of points A, B, and C, we get:

Area = 1/2 * |1(2*(-1) - 00) + 0(0(-1) - 00) + 0(00 - 2*(-1))| = 1/2 * |-2| = 1

Therefore, the area of triangle ABC is 1.