Consider a triangle △△𝐴𝐵𝐶ABC, whose co-ordinates are 𝐴(−3,3),𝐵(1,7)A(−3,3),B(1,7)and 𝐶(2,−2)C(2,−2). Let the point 𝑀M divides the line 𝐴𝐵AB in 1:3, the point 𝑁N divides the line 𝐴𝐶AC in 2:3 and the point O is the mid-point of 𝐵𝐶BC. Find out the area of triangle △△𝑀𝑁𝑂MNO (in sq. unit).

To find the area of triangle MNO, we first need to find the coordinates of points M, N, and O.

1. Point M divides AB in the ratio 1:3. So, the coordinates of M can be found using the formula for section formula in coordinate geometry:

   M = [(x1w2 + x2w1)/(w1+w2), (y1w2 + y2w1)/(w1+w2)]

   Substituting the given values, we get:

   M = [(-33 + 11)/4, (33 + 71)/4] = [-2, 3]

2. Similarly, point N divides AC in the ratio 2:3. So, the coordinates of N are:

   N = [(-33 + 22)/5, (33 + -22)/5] = [-1, 1]

3. Point O is the midpoint of BC. So, the coordinates of O are:

   O = [(x1 + x2)/2, (y1 + y2)/2]

   Substituting the given values, we get:

   O = [(1 + 2)/2, (7 - 2)/2] = [1.5, 2.5]

Now, we can find the area of triangle MNO using the formula for the area of a triangle given three points:

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

Substituting the coordinates of M, N, and O, we get:

Area = 1/2 * |-2(2.5 - 1) + -1(1 - 3) + 1.5(3 - 2.5)| = 1/2 * |-3 + 2 + 0.75| = 0.125 square units.