Question

The area of the quadrilateral formed by the points (2,1), (4,3), (-1,2), (-3,-2) is​

Answer 1

$Let \: A(2,1),B(4,3),C(-1,2) \:and \:D(-3,-2) \:are$

$Vertices \:of \: a \: Quadrilateral.$

$Join \: A\:and \: C$

$i)Area \: of \: \triangle ABC$

$=\frac{1}{2}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|$

$= \frac{1}{2}|2(3-2)+4(2-1)+(-1)(1-3)|$

$= \frac{1}{2}|2\times 1+4(1)+(-1)(-2)|$

$= \frac{1}{2}|2+4+2|$

$= \frac{8}{2}$

$= 4 \: --(1)$

$ii) Area \: of \: \triangle ACD$

$=\frac{1}{2}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|$

$= \frac{1}{2}|2(2+2)+(-1)(-2-1)+(-3)(1-2)|$

$= \frac{1}{2}|2\times 4+(-1)(-3)+(-3)(-1)|$

$= \frac{1}{2}|8+3+3|$

$= \frac{14}{2}$

$= 7\: --(2)$

$Now, iii) Area \:of \: Quadrilateral\:ABCD$

$= (1) + (2)$

$= 4+ 7$

$\green{= 11 \: square \:units}$

Therefore.,

$\red{ Area \:of \: Quadrilateral\:ABCD}$

$\green { = 11 \: square \:units}$

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