Given , x1=x y1 =y, x2=1,y2=2, x3=2, y3=1
area= 6 sq.units
area of triangle = 1/2 |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
6=1/2|x(2-1)+1(1-y)+2(y-2)|
12=|x+1-y+2y-4|
+or-12=x-3+y

12+3=x+y or -12+3=x+y
15=x+y or. x+y+9=0

hence proved.

Given information:
We are given the coordinates of three points A(x, y), B(1, 2), and C(2, 1). The area of triangle ABC is 6 square units.

Approach:
To solve this problem, we will use the formula for finding the area of a triangle when the coordinates of its vertices are known. The formula is as follows:

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

We will substitute the given coordinates into this formula and solve for the value of x and y.

Solution:

1. Let's substitute the given coordinates into the formula for finding the area of a triangle:

Area = 1/2 * |x(2 - 1) + 1(1 - y) + 2(y - 2)|

Simplifying further:

6 = 1/2 * |x + 1 - y + 2y - 4|

2. Multiply both sides of the equation by 2 to eliminate the fraction:

12 = |x + 1 - y + 2y - 4|

3. Remove the absolute value by considering both positive and negative cases:

Case 1: x + 1 - y + 2y - 4 = 12
Simplifying further:
x + y - 3 = 12
x + y = 15

Case 2: -(x + 1 - y + 2y - 4) = 12
Simplifying further:
-x - y + 3 = 12
-x - y = 9

4. Rearrange the equations to solve for x and y:

For Case 1: x + y = 15
y = 15 - x

For Case 2: -x - y = 9
y = -x - 9

5. Substitute the value of y from Case 1 into Case 2:

15 - x = -x - 9

6. Simplify and solve for x:

15 + 9 = -x + x
24 = 0

7. Since the equation 24 = 0 is not true, the solution x = 24 is not valid.

Conclusion:
From the given information, we have found that the possible solution is x + y = 15. Therefore, the equation xy = 15 or xy - 9 = 0 is valid.