Question

Find the area of the triangle with vertices at the points:
(i) (3, 8), (−4, 2) and (5, −1)
(ii) (2, 7), (1, 1) and (10, 8)
(iii) (−1, −8), (−2, −3) and (3, 2)
(iv) (0, 0), (6, 0) and (4, 3).

Answer 1

(i)
$$\begin{gathered} ∆=\frac{1}{2}\left|\begin{array}{ccc}3& 8& 1\\ -4& 2& 1\\ 5& -1& 1\end{array}\right| \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}3& 8& 1\\ -7& -6& 0\\ 5& -1& 1\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_1\right] \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}3& 8& 1\\ -7& -6& 0\\ 2& -9& 0\end{array}\right|\left[\mathrm{Applying}{R}_3\to R_3-R_1\right] \\ ∆=\frac{1}{2}\left|\begin{array}{cc}-7& -6\\ 2& -9\end{array}\right| \\ ∆=\frac{1}{2}\left(63+12\right) \\ ∆=\frac{1}{2}\left(75\right)=\frac{75}{2}\mathrm{square}\mathrm{units} \end{gathered}$$

(ii)
$$\begin{gathered} ∆=\frac{1}{2}\left|\begin{array}{ccc}2& 7& 1\\ 1& 1& 1\\ 10& 8& 1\end{array}\right| \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}2& 7& 1\\ -1& -6& 0\\ 10& 8& 1\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_1\right] \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}2& 7& 1\\ -1& -6& 0\\ 8& 1& 0\end{array}\right|\left[\mathrm{Applying}{R}_3\to R_3-R_1\right] \\ ∆=\frac{1}{2}\left|\begin{array}{cc}-1& -6\\ 8& 1\end{array}\right| \\ ∆=\frac{1}{2}\left(-1+48\right) \\ ∆=\frac{1}{2}\left(47\right)=\frac{47}{2}\mathrm{square}\mathrm{units} \end{gathered}$$

(iii)
$$\begin{gathered} ∆=\frac{1}{2}\left|\begin{array}{ccc}-1& -8& 1\\ -2& -3& 1\\ 3& 2& 1\end{array}\right| \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}-1& -8& 1\\ -1& 5& 0\\ 3& 2& 1\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_1\right] \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}-1& -8& 1\\ -1& 5& 0\\ 4& 10& 0\end{array}\right|\left[\mathrm{Applying}{R}_3\to R_3-R_1\right] \\ ∆=\frac{1}{2}\left|\begin{array}{cc}-1& 5\\ 4& 10\end{array}\right| \\ ∆=\frac{1}{2}\left|-10-20\right| \\ ∆=\frac{1}{2}\left(30\right)=15\mathrm{square}\mathrm{units} \end{gathered}$$

(iv)
$$\begin{gathered} ∆=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 6& 0& 1\\ 4& 3& 1\end{array}\right| \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 6& 0& 0\\ 4& 3& 1\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_1\right] \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 6& 0& 0\\ 4& 3& 0\end{array}\right|\left[\mathrm{Applying}{R}_3\to R_3-R_1\right] \\ ∆=\frac{1}{2}\left|\begin{array}{cc}6& 0\\ 4& 3\end{array}\right| \\ ∆=\frac{1}{2}\left(18-0\right) \\ ∆=\frac{1}{2}\left(18\right)=9\mathrm{square}\mathrm{units} \end{gathered}$$