If the points $A(x, y)$ , $B(1, 2)$ , and $C(2, 1)$ are vertices of a triangle whose area is $6\text{ sq. units}$ , show that: $x+y=15\quad \text{or}\quad x+y=-9$

Answer: $x+y=15\quad \text{or}\quad x+y=-9$

Step-by-step solution

Step 1: Recall the Area of a Triangle Formula

To find the area of a triangle given the coordinates of its vertices, we use a specific formula. This formula involves the coordinates of the three vertices $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ . The absolute value ensures that the area is always positive.

Step 2: Substitute the Given Coordinates and Area

We are given the vertices $A(x, y)$ , $B(1, 2)$ , and $C(2, 1)$ , and the area is $6$ square units. We substitute these values into the area formula. Here, $(x_1, y_1) = (x, y)$ , $(x_2, y_2) = (1, 2)$ , and $(x_3, y_3) = (2, 1)$ .

Step 3: Simplify the Expression

First, we multiply both sides by $2$ to remove the fraction. Then, we simplify the expression inside the absolute value by performing the multiplications and combining like terms. This gives us a simpler equation involving $x$ and $y$ .

Step 4: Solve for Two Possible Cases

Since the absolute value of an expression is $12$ , the expression itself can be either $12$ or $-12$ . This leads to two separate linear equations that we need to solve.

Step 5: Find the Final Equations

By isolating $x+y$ in each of the two equations, we arrive at the two possible relationships between $x$ and $y$ . These are the required conditions for the area of the triangle to be $6$ square units.

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If the points A(x,y)A(x, y)A(x,y), B(1,2)B(1, 2)B(1,2), and C(2,1)C(2, 1)C(2,1) are vertices of a triangle whose area is 6 sq. units6\text{ sq. units}6 sq. units, show that: x+y=15orx+y=−9x+y=15\quad \text{or}\quad x+y=-9x+y=15orx+y=−9