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If (x(1),x(2))^(2)+(y(1)-y(2))^(2)=a^(2)...

If `(x_(1),x_(2))^(2)+(y_(1)-y_(2))^(2)=a^(2), (x_(2)-x_(3))^(2)+(y_(2)-y_(3))^(2)=b^(2) (x_(3)-x_(1))^(2)+(y_(3)-y_(1))^(2)=c^(2).` where a,b,c are positive then prove that `4 |{:(x_(1),,y_(1),,1),(x_(2) ,,y_(2),,1),( x_(3),, y_(3),,1):}| = (a+b+c) (b+c-a)
(c+a-b)(a+b-c)`

Text Solution

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we have `sqrt((x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2))=a`
`sqrt((x^(2)-x_(3))^(2)+(y_(2)-y_(3))^(2))=b`
`sqrt((x_(3)-x_(1))^(2)+(y_(3)-y_(1))^(2))=c`
Consider triangle having vertices `A(x_(1),y_(1)),B(x_(2),y_(2))` and
So , `AB=a,BC=b " and "AC= c.Also ,2s=a+b+c` where s is semi-perimetre.
Area of triangle ABC
`Delta=(1)/(2) ||{:(x_(1),,y_(1),,1),(x_(2),,y_(2),,1),(x_(3),,y_(3),,1):}||`
Also `Delta= sqrt(s(s-a)(s-b)(s-c))`
`:. ||{:(x_(1),,y_(1),,1),(x_(2),,y_(2),,1),(x_(3),,y_(3),,1):}||=sqrt(s(s-a)(s-b)(s-c))`
Squaring and simplifying we get
`4 |{:(x_(1),,y_(1),,1),(x_(2),,y_(2),,1),(x_(3),,y_(3),,1):}|^(2)`
=`(a+b+c)(b+c-a)(c+a-b)(a+b-c)`
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