If the points $A(x, y)$, $B(2, 4)$, and $C(4, 2)$ are vertices of a triangle whose area is $9\text{ sq. units}$, show that: $x+y=15\quad \text{or}\quad x+y=-3$

Step 1: Recall the Area of a Triangle Formula

To find the area of a triangle with given vertices, we use the determinant formula. The absolute value ensures the area is always positive. We are given the vertices $A(x, y)$, $B(2, 4)$, and $C(4, 2)$, and the area is $9$ square units.

Step 2: Substitute the Given Values into the Formula

Now, we substitute the coordinates of the points $A(x_1, y_1) = (x, y)$, $B(x_2, y_2) = (2, 4)$, and $C(x_3, y_3) = (4, 2)$ into the area formula. We also set the area equal to $9$.

Step 3: Simplify the Expression

We multiply both sides by $2$ to remove the fraction and then simplify the expression inside the absolute value by combining like terms. This gives us a simplified equation involving $x$ and $y$.

Step 4: Solve for Two Possible Cases

Since the absolute value of an expression is $18$, the expression itself can be either $18$ or $-18$. This leads to two separate linear equations.

Step 5: Solve the First Case

For the first case, we add $12$ to both sides and then divide the entire equation by $2$ to get the first relationship between $x$ and $y$.

Step 6: Solve the Second Case

For the second case, we again add $12$ to both sides and then divide by $2$ to obtain the second relationship between $x$ and $y$. This completes the proof.

More from this chapter

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