Question

Find area of triangle with vertices at the point given in each of the following
(i)$(1,0),(6,0),(4,3)$
(ii)$(2,7),(1,1),(10,8)$
(iii)$(-2,-3),(3,2),(-1,-8)$

Answer 1

Area of triangle having vertices as $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given as

$=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

(i) Area of triangle $=\frac{1}{2}\left|\begin{array}{ccc}1& 0& 1\\ 6& 0& 1\\ 4& 3& 1\end{array}\right|$

Expanding along second column,

$\Delta =\frac{1}{2}[-3\left|\begin{array}{cc}1& 1\\ 6& 1\end{array}\right|]$

$=\frac{1}{2}[-3(1-6\left)\right]$

$\Rightarrow \Delta =\frac{15}{2}sq.units$

(ii) Area of triangle $=\frac{1}{2}\left|\begin{array}{ccc}2& 7& 1\\ 1& 1& 1\\ 10& 8& 1\end{array}\right|$

$\Delta =\frac{1}{2}\left[2\right(1-8)-7(1-10)+1(8-10\left)\right]$

$=\frac{1}{2}[-14+63-2]$

$\Delta =\frac{1}{2}47sq.units$

(iii) Area of triangle $=\frac{1}{2}\left|\begin{array}{ccc}-2& -3& 1\\ 3& 2& 1\\ -1& -8& 1\end{array}\right|$

$\Delta =\frac{1}{2}[-2(2+8)+3(3+1)+1(-24+2\left)\right]$

$=\frac{1}{2}\left|\right[-20+12-22\left]\right|$

$\Delta =|\frac{-30}{2}|$