91Ó°ÊÓ

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 $\left(a,b\right)=\left(5,4\right)$, $\left(c,d\right)=\left(-3,-2\right)$and $\left(e,f\right)=\left(3,-4\right)$. So, substitute 5 for a, 4 for b, -3 for c, -2 for d, 3 for e and -4 for f into the expression $A=\frac{1}{2}\left|\begin{array}{ccc}a& b& 1\\ c& d& 1\\ e& f& 1\end{array}\right|$.

$A=\frac{1}{2}\left|\begin{array}{ccc}5& 4& 1\\ −3& −2& 1\\ 3& −4& 1\end{array}\right|$

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

$\begin{array}{c}A=\frac{1}{2}\left|\begin{array}{ccc}5& 4& 1\\ −3& −2& 1\\ 3& −4& 1\end{array}\right|\\ =\frac{1}{2}\left(5\left|\begin{array}{cc}−2& 1\\ −4& 1\end{array}\right|−4\left|\begin{array}{cc}−3& 1\\ 3& 1\end{array}\right|+\left|\begin{array}{cc}−3& −2\\ 3& −4\end{array}\right|\right)\end{array}$

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 $2×2$determinant.

localid="1648217738738" $\begin{array}{c}A=\frac{1}{2}\left(5\left[−2−\left(−4\right)\right]−4\left[−3−3\right]+\left[−3\left(−4\right)−3\left(−2\right)\right]\right)\\ =\frac{1}{2}\left(5\left[2\right]−4\left[−6\right]+\left[12+6\right]\right)\\ =\frac{1}{2}\left(10+24+18\right)\\ =\frac{1}{2}×52\\ =26\end{array}$

Therefore, area of the triangle is 26 square units.