The incentre of the triangle, whos l(-36, 7). (20, 7) and (0. 8) is

1.	The incentre of the triangle, whos l(-36, 7). (20, 7) and (0. 8) is
Answer» 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). HIT THE LIKE BUTTON IF THIS HELPED YOU!

Answer»

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).

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