find the incentre of the triangle whose vertices are - 36 ,7 20,7 and 0, 8
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Step-by-step explanation:
Let A=(−36,7) B=(20,7) and C=(0,−8)
∴a=BC=
(0−20)
2
+(−8−7)
2
=
400+225
=
625
=25
b=CA=
(−36−0)
2
+(7+8)
2
=
1296+225
=
1521
=39
c=AB=
(20+36)
2
+(7−7)
2
=
3136+0
=
3136
=56
Let I(x,y) be the in-centre of △ABC
∴x=
a+b+c
ax
1
+bx
2
+cx
3
and y=
a+b+c
ay
1
+by
2
+cy
3
Substituting the above values in the above formula we get
x=
25+39+56
25×−36+39×20+56×0
=
120
−120
=−1
and y=
25+39+56
25×7+39×7+56×−8
=
120
0
=0
Thus, the in-centre is at (−1,0)
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