Question

the incentre of the triangle formed by lines x+y=1,x=1,y=1is
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Answer 1

Answer:

( 1/√2, 1/√2 )

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Step-by-step explanation:

Let the incentre be at ( a, b ).

By symmetry, the centre must be on the line x = y.

So its coordinates are ( a, a ).

The distance from ( a, a ) to the line x + y = 1 is just the distance from ( a, a ) to the point ( 1/2, 1/2 ).

The distance from the line x = 1 is 1 - a.

At the incentre, these distances must be the same.  So...

square of distance from ( 1/2, 1/2 ) = square of distance from line x = 1

=> ( a - 1/2 )² + ( a - 1/2 )² = ( 1 - a )²

=> 2 ( a - 1/2 )² = ( 1 - a )²

=> 2a² - 2a + 1/2 = a² - 2a + 1

=> a² = 1/2

=> a = ± 1 / √2.

As the incentre is inside the triangle, we must take the positive value.

So the incentre is at ( 1/√2, 1/√2 ).

[ Incidentally, because of how we got the equation in the first place, the negative option tells us that the point ( -1/√2, -1/√2 ) is one of the excentres. ]