InterviewSolution
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InterviewSolution
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The x-coordinate of the incentre of the trianglethat has the coordinates of mid points of its sides as (0, 1), (1, 1) and (1,0) is(1) `2-sqrt(2)`(2) `1+sqrt(2)`(3) `1-sqrt(2)`(4) `2+sqrt(2)` |
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Answer» `0 = (a+x)/2` `x=-a` `1= (b+y)/2` `y= 2-b` now,` 1= (-a + 2 -a)/2` `2 = -2a + 2` `a= 0` & `4-2b = 0` `b=2` `AC= sqrt((0-2)^2 + (2-0)^2) = sqrt 8 ` `I_n = (ax_1 + bx_2 + c x_3)/(a+b+c)` where, a,b,c are length of the sides & `x_1, x_2 , x_3` are opposite vertex coordinates so,` (2 xx 2 + 2 xx 0 + sqrt8 xx 0)/(2 + 2 + sqrt8)` `= 4/(4 + 2 sqrt2) xx (4 - 2sqrt 2)/(4 - 2sqrt 2)` `= (16- 8 sqrt2)/(16- 8)` `= (16- 8 sqrt2)/8` `= (4-2sqrt2)/2 = 2 - sqrt2` option 1 is correct |
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