Points \(A(7,12)\), \(B(3,2)\), \(C(11,2)\), \(D(6,2)\) and \(E(10,2)\) are plotted on the Cartesian plane. The point \(F(8,k)\) lies inside \(\triangle ABC\) so that the area inside of \(\triangle ABC\) but outside of \(\triangle DEF\) is equal to \(32\text{ units}^2\).
Determine the value of \(k\), the \(y\)-coordinate of \(F\).
We will calculate the area of \(\triangle ABC\), calculate the area of \(\triangle DEF\), and then use the given information that the difference between these areas is equal to \(32\text{ units}^2\).
We will use the fact that the distance between two points that have the same \(x\)-coordinate is the positive difference between their \(y\)-coordinates. We will also use the fact that the distance between two points that have the same \(y\)-coordinate is the positive difference between their \(x\)-coordinates.
Since \(B\) and \(C\) both have \(y\)-coordinate \(2\), \(BC\) is a horizontal line. Thus, \(BC = 11-3 = 8\).
In \(\triangle ABC\), drop a perpendicular from vertex \(A\) to \(M\) on \(BC\). Since \(BC\) is horizontal, then \(AM\) is vertical. Since every point on a vertical line has the same \(x\)-coordinate, \(M\) has \(x\)-coordinate \(7\). Similarly, since \(M\) is on the horizontal line through \(B(3,2)\) and \(C(11,2)\), \(M\) has \(y\)-coordinate \(2\). Therefore, \(AM = 12-2 = 10\).
Thus, the area of \(\triangle ABC\) is equal to \(\dfrac{BC\times AM}{2}=\dfrac{8\times 10}{2}=40\).
Since \(D\) and \(E\) both have \(y\)-coordinate \(2\), \(DE\) is a horizontal line. Thus, \(DE = 10-6 = 4\).
In \(\triangle DEF\), drop a perpendicular from vertex \(F\) to \(N\) on \(DE\). Since \(DE\) is horizontal, then \(FN\) is vertical. Since every point on a vertical line has the same \(x\)-coordinate, \(N\) has \(x\)-coordinate \(8\). Similarly, since \(N\) is on the horizontal line through \(D(6,2)\) and \(E(10,2)\), \(N\) has \(y\)-coordinate \(2\). Therefore, \(FN = k-2\).
Thus, the area of \(\triangle DEF\) is equal to \(\dfrac{DE\times FN}{2}=\dfrac{4\times (k-2)}{2}=2(k-2)\).
We now use the given information that the area inside of \(\triangle ABC\) but outside of \(\triangle DEF\) is equal to \(32\text{ units}^2\) to obtain the equation \(40-2(k-2)=32\).
Subtracting \(32\) from each side, we have \(8-2(k-2) =0\).
Adding \(2(k-2)\) to each side, we have \(8 =2(k-2)\), which simplifies to \(4=k-2\), or \(k=6\).
Therefore, the value of \(k\), the \(y\)-coordinate of point \(F\), is \(6\).