Find area of the triangle with vertices at the point given in each of the following:
(i)$\left( {1,0} \right),\left( {6,0} \right),\left( {4,3} \right)$
(ii)$\left( {2,7} \right),\left( {1,1} \right),\left( {10,8} \right)$
(iii)$\left( { - 2, - 3} \right),\left( {3,2} \right),\left( { - 1, - 8} \right)$
(i) Area of triangle $= \cfrac{1}{2}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&1\\6&0&1\\4&3&1\end{array}} \right|$
$= \cfrac{1}{2}\left[ {1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1\\3&1\end{array}} \right| - 0 + 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6&0\\4&3\end{array}} \right|} \right]$
[Expanding along ${R_1}$]
$= \cfrac{1}{2}[1(0 - 3) + 1(18 - 0)] = \cfrac{{15}}{2} = 7.5$ sq. units.
(ii) Area of triangle $= \cfrac{1}{2}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&7&1\\1&1&1\\{10}&8&1\end{array}} \right|$
$= \cfrac{1}{2}\left[ {2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\8&1\end{array}} \right| - 7\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\{10}&1\end{array}} \right| + 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1\\{10}&8\end{array}} \right|} \right]$
[Expanding along ${R_1}$]
$= \cfrac{1}{2}[2(1 - 8) - 7(1 - 10) + 1(8 - 10)]$
$= \cfrac{1}{2}[ - 14 + 63 - 2] = \cfrac{{47}}{2} = 23.5$ sq. units
(iii) Area of triangle $= \cfrac{1}{2}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}&{ - 3}&1\\3&2&1\\{ - 1}&{ - 8}&1\end{array}} \right|$
$= \cfrac{1}{2}\left[ { - 2\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&1\\{ - 8}&1\end{array}} \right| + 3\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&1\end{array}} \right| + 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\{ - 1}&{ - 8}\end{array}} \right|} \right]$
[Expanding along${R_1}$]
$= \cfrac{1}{2}[ - 2(2 + 8) + 3(3 + 1) + 1( - 24 + 2)]$
$= \cfrac{1}{2}[ - 20 + 12 - 22] = \cfrac{{ - 30}}{2} = - 15$
$\therefore$ Area $=$ 15 square units. [Absolute value]
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Determinants. Curated by Sachin Sharma. Free for all students.