find the incentre of the triangle formed by the points
Answers
Answer:
Coordinates of the In center is ( 1.7 , 1.8 ).
Step-by-step explanation:
Given: Coordinates of the triangle A( 7 , 9 ) , B( 3 , -7 ) and C( -3 , 3 )
To find: In center of the triangle.
In center of the triangle = (\frac{a\,A_x+b\,B_x+c\,C_x}{p},\frac{a\,A_y+b\,B_y+c\,C_y}{p})(
p
aA
x
+bB
x
+cC
x
,
p
aA
y
+bB
y
+cC
y
)
where, A_x\,,\,B_x\,,\,C_xA
x
,B
x
,C
x
are x coordinates of the vertex.
A_y\,,\,B_y\,,\,C_yA
y
,B
y
,C
y
are y coordinates of the vertex.
a , b , c are the length of sides opposite to vertex A , B and C respectively.
p is the perimeter of the triangle.
length of the side AB opposite to vertex C , c = \sqrt{(3-7)^+(-7-9)^2}=\sqrt{16+256}=16.5
(3−7)
+
(−7−9)
2
=
16+256
=16.5
length of the side CB opposite to vertex A , a = \sqrt{(-3-3)^+(3-(-7))^2}=\sqrt{36+100}=11.7
(−3−3)
+
(3−(−7))
2
=
36+100
=11.7
length of the side AC opposite to vertex B , b = \sqrt{(-3-7)^+(3-9)^2}=\sqrt{100+36}=11.7
(−3−7)
+
(3−9)
2
=
100+36
=11.7
p = 11.7 + 11.7 + 16.5 = 39.9
So, In center
(\frac{11.7(7)+11.7(3)+16.5(-3)}{39.9},\frac{11.7(9)+11.7(-7)+16.5(3)}{39.9})=(\frac{81.9+35.1-49.5}{39.9},\frac{105.3-81.9+49.5}{39.69})(
39.9
11.7(7)+11.7(3)+16.5(−3)
,
39.9
11.7(9)+11.7(−7)+16.5(3)
)=(
39.9
81.9+35.1−49.5
,
39.69
105.3−81.9+49.5
)
=(\frac{67.5}{39.9},\frac{72.9}{39.9})=(1.7,1.8)=(
39.9
67.5
,
39.9
72.9
)=(1.7,1.8)
Therefore, Coordinates of the In center is ( 1.7 , 1.8 )
I wish it helps u frnd........
