Q14. - Algebra 2

Question

Find the area of the triangle shown at the right.

Step-by-Step Solution

Answer

Area of the triangle is 26 square units.

Step 1 - Define area of triangle

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 $A=\frac{1}{2}\left|\begin{array}{ccc}a& b& 1\\ c& d& 1\\ e& f& 1\end{array}\right|$.

 

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

Step 2 - Find the determinant

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

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

Step 3 - Evaluate 2 × 2 determinants

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.

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

Therefore, area of the triangle is 26 square units.