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

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(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

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(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.