Question

Find the co-ordinates of the in-centre of the triangle whose vertices are (-36,7),(20,7)and (0,-8).
Please answer with explanation

Answer 1

Given, vertices of the triangle are A (-36,7), B (20,7) and C (0,-8)

a = BC = root (0-20)^2 + (-8-7)^2

            = root(-20)^2 + (-15)^2

           = root 400 +  625

          = root 525

          = 25.


b = CA = root (36-0)^2 + (7-(-8))^2

            = root 1296 + 225

           = root 1521

           = 39.


c = AB = root 20-(-36)^2 + (7-7)^2

            = root (20 + 36)^2 + 0

           = root 56^2

           = 56.

a = 25, b=39,c=56.

We now that incentre of the triangle is 

(ax1 + bx2 + cx3)/(a+b+c),(ay1 + by2 + cy3)/(a+b+c)    ------------ (2)

Subsitute values in (2), we get

(25(-36)+39(20)+56(0)/(25+39+56),25(7)+39(7)+56(-8)/(25+39+56)) 

= ((-120)/120,(448-448)/120)

= (-1,0).

Therefore the incentre is (-1,0).

Answer 2

Step-by-step explanation:

A] Given A(7, -36) B(7,20) C(-8,0)

a = BC =

(−8−7)

2

+(0−20)

2

=

400+625

=25

b = CA =

(−8−7)

2

+(0+36)

2

=39

c = AB =

(7−7)

2

+((20+36)

2

)

=56

a = 25, b = 39, c = 56

In center =(

x+b

ax

1

+bx

2

+cx

3

),(

a+b+c

ay

1

+by

2

+cy

3

)

=(

25+39+56

25×7+36×7+56×8

),(

25+39+56

25×−36+39×20+56×0

)

=(

120

0

,

120

−120

)=(0,−1)

∴ In center is (0,-1)