Question

3s=a+b+c then (s-a)^2 + (s-b)^2 + (s-c)^2 =?​

Answer 1

SOLUTION

GIVEN

$\sf{3s = a + b + c \: }$

TO DETERMINE

$\sf{ {(s - a)}^{2} + {(s - b)}^{2} + {(s - c)}^{2} }$

EVALUATION

Here it is given that

$\sf{3s = a + b + c \: }$

$\sf{Let \: \: x = s - a, y = s - b, z = s - c}$

Then

$\sf{x + y + z \: }$

$= \sf{s - a + s - b + s - c \: }$

$\sf{ = 3s - a - b - c}$

$\sf{ = 3s - (a + b + c)}$

$\sf{ = 3s - 3s}$

$\sf{ = 0}$

Now we are aware of the identity that

$\sf{ {(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx)}$

$\implies \sf{ {(0)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx)}$

$\implies \sf{ {x}^{2} + {y}^{2} + {z}^{2} = - 2(xy + yz + zx)}$

$\implies \sf{ {(s - a)}^{2} + {(s - b)}^{2} + {(s - c)}^{2} = - 2{(s - a)} {(s - b)} {(s - c)} }$

FINAL ANSWER

$\sf{ {(s - a)}^{2} + {(s - b)}^{2} + {(s - c)}^{2} = - 2{(s - a)} {(s - b)} {(s - c)} }$

━━━━━━━━━━━━━━━━

LEARN MORE FROM BRAINLY

If the distance of the point P from the point

(3, 4) is √10 and abscissa of P is double its ordinate, find the co-ordinate

https://brainly.in/question/24239320