find the incentre of the triangle formed by lines x=1,y=1,x+y=1and
a=1,b=√2,c=1
Answers
Answer:
The incentre of the triangle formed by the lines x= 1, y=1 and x + y = 1 is given by,
Consider the attached figure, while going through the following steps.
Incenter of the triangle is given by the formula,
I = (\dfrac{ax_1+bx_2+cx_3}{a+b+c}, \dfrac{ay_1+by_2+cy_3}{a+b+c})I=(
a+b+c
ax
1
+bx
2
+cx
3
,
a+b+c
ay
1
+by
2
+cy
3
)
The given lines, x = 1, y = 1 and x + y = 1 intersect each other at points A (1, 1), B (0, 1) and C (1, 0).
we, have,
a = BC = √2
b = AC = 1
c = AB = 1
Therefore, the incenter is given by,
I = (\dfrac{\sqrt 2 (1) +1(0) + 1 (1)}{\sqrt 2 +1+1}, \dfrac{\sqrt 2 (1) +1 (1) + 1 (0)}{\sqrt 2 +1+1})I=(
2
+1+1
2
(1)+1(0)+1(1)
,
2
+1+1
2
(1)+1(1)+1(0)
)
I = (\dfrac{\sqrt 2+1}{\sqrt 2 +2}, \dfrac{\sqrt 2 +1 }{\sqrt 2 +2})I=(
2
+2
2
+1
,
2
+2
2
+1
)
Upon rationalizing the above terms, we get,
I = (\dfrac{1}{\sqrt 2}, \dfrac{1}{\sqrt 2})I=(
2
1
,
2
1
)