InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Find the coordinates of the center of the circle inscribed in a triangle whose vertices are (-36,7) , (20,7) and (0,-8). |
|
Answer» Let `ABC` is the triangle with the vertices `A(-36,7), B(20,7) and C(0,-8)`. Then, `AB = sqrt((20-(-36))^2 + (7-7)^2) = 56` `BC = sqrt((0-(-20))^2+((-8)-7)^2 ) )= sqrt(400+225) = sqrt(625) = 25` `AC = sqrt((-36-0)^2 + (7-(-8))^2)) = sqrt(1521) = 39` Let, `O(O_x,O_y)` is the center of the circle inscribed in triangle `ABc`. Then, `O_x = (A_xa+B_xb+C_xc)/P and O_y = (A_ya+B_yb+C_yc)/P` Here, `P = AB+BC+AC = 56+25+39 = 10` and `a = 25,b = 39, c = 56` So, `O_x = (-36**25+20**39+0**56)/120 = -120/120 = -1` `O_y = (7**25+7**39+(-8)**56)/120 = 0/120 = 0` So, coordinates of the center of the circle are `O(-1,0)`. |
|