Question

Find area of the 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

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

$\triangle$=\frac{1}{2}$$\begin{vmatrix}1&0&1\\\6&0&1\\\4&3&1\end{vmatrix}

=$\frac{1}{2}$[1(0-3)-0(6-4)+1(18-0)]

=$\frac{1}{2}$[-3+18]=$\frac{15}{2}$ square units

---

(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

△=\frac{1}{2}$$\begin{vmatrix}2&7&1\\\1&1&1\\\10&8&1\end{vmatrix}

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

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

=$\frac{1}{2}$[-16+63]

=$\frac{47}{2}$ square units

---

(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8)
is given by the relation,

△=\frac{1}{2}$$\begin{vmatrix}-2&-3&1\\\3&2&1\\\\-1&-8&1\end{vmatrix}

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

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

=-$\frac{30}{2}$=-15

Hence, the area of the triangle is I-15I=15 square units.