GEOMETRY Find the value of x such that the area of a triangle whose vertices have coordinates (6, 5), (8, 2) and (x, 11) is 15 square units.

The area of triangle having vertices at $\left(a,b\right)$, $\left(c,d\right)$and $\left(e,f\right)$is $\left|A\right|$, where $A=\frac{1}{2}\left|\begin{array}{ccc}a& b& 1\\ c& d& 1\\ e& f& 1\end{array}\right|$.

Here vertices of the triangle are $\left(a,b\right)=\left(6,5\right)$, $\left(c,d\right)=\left(8,2\right)$and $\left(e,f\right)=\left(x,11\right)$. So, substitute 6 for a, 5 for b, 8 for c, 2 for d, x for e and 11 for f into the expression $A=\frac{1}{2}\left|\begin{array}{ccc}a& b& 1\\ c& d& 1\\ e& f& 1\end{array}\right|$.

$A=\frac{1}{2}\left|\begin{array}{ccc}6& 5& 1\\ 8& 2& 1\\ x& 11& 1\end{array}\right|$

Find the determinant in this case by using expansion by minors.

$\begin{array}{c}A=\frac{1}{2}\left|\begin{array}{ccc}6& 5& 1\\ 8& 2& 1\\ x& 11& 1\end{array}\right|\\ =\frac{1}{2}\left(6\left|\begin{array}{cc}2& 1\\ 11& 1\end{array}\right|-5\left|\begin{array}{cc}8& 1\\ x& 1\end{array}\right|+\left|\begin{array}{cc}8& 2\\ x& 11\end{array}\right|\right)\end{array}$

The determinant of second order matrix is found by calculating the difference of the product of the two diagonals, that is., $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$.Apply this definition to find the $2\times 2$determinant.

$\begin{array}{c}A=\frac{1}{2}\left(6\left[2-11\right]-5\left[8-x\right]+\left[8\left(11\right)-2x\right]\right)\\ =\frac{1}{2}\left(6\left[-9\right]-5\left(8\right)-5\left(-x\right)+\left[88-2x\right]\right)\\ =\frac{1}{2}\left(-54-40+5x+88-2x\right)\\ =\frac{1}{2}\left(3x-6\right)\end{array}$

Since area of triangle is given to be 15 square units, therefore, $\frac{1}{2}\left(3x-6\right)=15$. Solve the equation for x.

$\begin{array}{c}\frac{1}{2}\left(3x-6\right)=15\\ 3x-6=30\\ 3x-6+6=30+6\\ 3x=36\\ x=12\end{array}$

Therefore, value of $x=12$.