Question

The area of the triangle formed by the vertices (a 1/a), (b, 1/b) and (c, 1/ c) is

Answer 1

Answer:

answer for the given problem is given

Answer 2

$Let \: A(a,\frac{1}{a}) = ( x_{1} , y_{1}) ,$

$B(b,\frac{1}{b}) = ( x_{2} , y_{2}) \:and$

$C(c,\frac{1}{c}) = ( x_{3} , y_{3}) \:are \: vertices$

$of \:a \: triangle .$

$Area \: of \: the \: triangle$

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

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

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

$= \frac{1}{2}| \frac{ac-ab}{bc} + \frac{ab-bc}{ac}+ \frac{bc-ac}{ab} |$

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

$= \frac{1}{2}| \frac{a^{2}c-a^{2}b+ab^{2}-b^{2}c+bc^{2}-ac^{2}}{abc}|$

Therefore.,

$\red{Area \: of \: the \: triangle}$

$\green { = \frac{1}{2}| \frac{a^{2}c-a^{2}b+ab^{2}-b^{2}c+bc^{2}-ac^{2}}{abc}| }$

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