Find the area of the triangle whose vertices are (3,1),)(1,3) and (-3,-2).

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written 5.6 years ago by

teamques10 ★ 69k

modified 3.0 years ago by

pedsangini276 • 4.8k

engineering mathematics

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1 Answer

1.0kviews

written 5.6 years ago by

teamques10 ★ 69k

modified 2.2 years ago by

samarthmhask439 • 30

Solution:

$let (x_{1},y_{1} = (-3,1),(x_{2},y_{2}) = (1,-3) and (x_{2},y_{2}) = (2,3)$

$A = \frac{1}{2} \begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}$

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

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

A = 14 sq.units

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