Find the area of the triangle with vertices at the points:

Solution:

(i)

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & -1 & 1\end{array}\right|$

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}3 & 8 & 1 \\ -7 & -6 & 0 \\ 5 & -1 & 1\end{array}\right| \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\right]$

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}3 & 8 & 1 \\ -7 & -6 & 0 \\ 2 & -9 & 0\end{array}\right| \quad\left[\right.$ Applying $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$

$\Delta=\frac{1}{2}\left|\begin{array}{cc}-7 & -6 \\ 2 & -9\end{array}\right|$

$\Delta=\frac{1}{2}(63+12)$

$\Delta=\frac{1}{2}(75)=\frac{75}{2}$ square units

(ii)

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1\end{array}\right|$

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}2 & 7 & 1 \\ -1 & -6 & 0 \\ 10 & 8 & 1\end{array}\right| \quad$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ ]

$\Delta=\frac{1}{2}\left|\begin{array}{rrr}2 & 7 & 1 \\ -1 & -6 & 0 \\ 8 & 1 & 0\end{array}\right| \quad\left[\right.$ Applying $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$

$\Delta=\frac{1}{2}\left|\begin{array}{cc}-1 & -6 \\ 8 & 1\end{array}\right|$

$\Delta=\frac{1}{2}(-1+48)$

$\Delta=\frac{1}{2}(47)=\frac{47}{2}$ square units

(iii)

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}-1 & -8 & 1 \\ -2 & -3 & 1 \\ 3 & 2 & 1\end{array}\right|$

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}-1 & -8 & 1 \\ -1 & 5 & 0 \\ 3 & 2 & 1\end{array}\right| \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\right]$

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}-1 & -8 & 1 \\ -1 & 5 & 0 \\ 4 & 10 & 0\end{array}\right| \quad$ [Applying $R_{3} \rightarrow R_{3}-R_{1}$ ]

$\Delta=\frac{1}{2}\left|\begin{array}{cc}-1 & 5 \\ 4 & 10\end{array}\right|$

$\Delta=\frac{1}{2}|-10-20|$

$\Delta=\frac{1}{2}(30)=15$ square units

(iv)

$\Delta=\frac{1}{2}\left|\begin{array}{lll}0 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1\end{array}\right|$

$\Delta=\frac{1}{2}\left|\begin{array}{lll}0 & 0 & 1 \\ 6 & 0 & 0 \\ 4 & 3 & 1\end{array}\right| \quad$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ ]

$\Delta=\frac{1}{2}\left|\begin{array}{lll}0 & 0 & 1 \\ 6 & 0 & 0 \\ 4 & 3 & 0\end{array}\right| \quad$ [Applying $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$

$\Delta=\frac{1}{2}\left|\begin{array}{ll}6 & 0 \\ 4 & 3\end{array}\right|$

$\Delta=\frac{1}{2}(18-0)$ $\Delta=\frac{1}{2}(18)=9$ square units