# Find area of the triangle with vertices at the point given in each of the following:

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) $(-2,-3),(3,2),(-1,-8)$

Solution:

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

\begin{aligned} \Delta &=\frac{1}{2}\left|\begin{array}{lll}1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1\end{array}\right| \\ &=\frac{1}{2}[1(0-3)-0(6-4)+1(18-0)] \\ &=\frac{1}{2}[-3+18]=\frac{15}{2} \text { square units } \end{aligned}

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

\begin{aligned} \Delta &=\frac{1}{2}\left|\begin{array}{lll}2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1\end{array}\right| \\ &=\frac{1}{2}[2(1-8)-7(1-10)+1(8-10)] \\ &=\frac{1}{2}[2(-7)-7(-9)+1(-2)] \\ &=\frac{1}{2}[-14+63-2]=\frac{1}{2}[-16+63] \\ &=\frac{47}{2} \text { square units } \end{aligned}

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

is given by the relation,

$\Delta=\frac{1}{2}\left|\begin{array}{rrr}-2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1\end{array}\right|$

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

$=\frac{1}{2}[-2(10)+3(4)+1(-22)]$

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

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

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