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

Answer:

a = BC = √(0-20)² + (-8-7)²

          =  √(-20)² + (-15)²

          =  √400+225

          =  √625

          =  √25 x 25

          =  25  

b = CA = √(36-0))² + (7-(-8))²

          =  √(36)² + (7+8)²

          =  √1296 + 15²

          =  √1296+225

          =  √1521

          =  √39 x 39

          =  39  

c = AB = √(20-(-36))² + (7-7)²

          =  √(20+36)² + (0)²

          =  √56²

          =   √56 x 56

           =  56

Incentre I of the triangle is

[(ax₁ + bx₂ + cx₃)/(a+b+c),(ay₁ + by₂ + cy₃)/(a+b+c)]

x₁ = -36 y₁ = 7 x₂ = 20 y₂ = 7  x₃ = 0 y₃ = -8

         a = 25 b = 39  c = 56

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

= [(-900+780+0)/(120),(175+273-448/(120)]

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

= (-1,0)

So the incentre is (-1,0)

Answer 2

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