Question

find the area of a triangle where coordinates of vertices are ( a , b + c ) , ( b , c + a ) and ( c , a + b )​

Answer 1

Answer:

a+b+c =180and b+c+a =180

Answer 2

Answer:

$\red{ Area \:of \:the \: triangle }\green {=0}$

Step-by-step explanation:

$Let \:A(a,b+c) = (x_{1},y_{1}) , \\(b,c+a) = (x_{2},y_{2}) ,\:and C(c,a+b) = (x_{3},y_{3}) \:are\\vertices \:of \: triangle \:ABC$

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

$=\frac{1}{2} | a[c+a-(a+b)]+b[a+b-(b+c)] +c[b+c-(c+a)]|$

$= \frac{1}{2} | a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)|$

$= \frac{1}{2}| a(c-b)+b(a-c)+c(b-a)| \\= \frac{1}{2}|ac-ab+ab-bc+bc-ac|\\=0$

$A,B \:and \:C \:are \: collinear .$

Therefore.,

$\red{ Area \:of \:the \: triangle }\green {=0}$

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