Question

Prove that A(1,1) B(3,2) C(5,3) are not the vertices of triangle ABC

Answer 1

$Given,\: A(x_{1},y_{1}) = ( 1,1 ) , \\B(x_{2},y_{2}) = ( 3,2) , C(x_{3},y_{3}) = ( 5,3 )$

/* We have to show that A,B and C are collinear points */

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

$= \frac{1}{2}|1(2-3)+3(3-1)+5(1-2)|\\= \frac{1}{2}|-1+3\times 2 +5(-1)|\\= \frac{1}{2}| -1+6-5|\\= \frac{1}{2} \times 0 \\= 0$

$Area \: of \: \triangle ABC = 0$

Therefore.,

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

$A,B \:and \: C \: are \: doesn't \:form \:a \\triangle .$

•••♪

Answer 2

$\huge{ \mathtt{ \fbox{Solution :)}}}$

Given ,

The three points are A(1,1) , B(3,2) and C(5,3)

We know that ,

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

Thus ,

$\sf \hookrightarrow Area </p><p> = \frac{1}{2}|1(2-3)+3(3-1)+5(1-2)|\\ \\ \sf \hookrightarrow Area= \frac{1}{2}|-1+3\times 2 +5(-1)|\\ \\ \sf \hookrightarrow Area= \frac{1}{2}| -1+6-5|\\ \\ \sf \hookrightarrow Area= \frac{1}{2} \times 0 \\ \\ \sf \hookrightarrow Area= 0 \: \: {units}^{2}$

Since , if area of triangle is 0 square units , then its vertices will be collinear

Hence , A(1,1) , B(3,2) and C(5,3) are not the vertices of triangle

_____________ Keep Smiling ☺