Question

Find the incenter of the triangle formed by the following straight lines
(i) x + 1 = 0, 3x - 4y = 5 and 5x + 12y = 27
(ii) x + y - 7 = 0, x - y + 1 = 0 and x - 3y + 5 = 0​

Answer 1

Step-by-step explanation:

Let AB be x+y = 7

BC be x-y +1 = 0

AC be x-3y +5 = 0

A is 4 ,3

B is 3, 4

C is 1 ,2

AB = c = √2

AC = b = √10

BC = a = 2√2

16A^2 = 4*10*8-( 2–10–8)^2 = 320–256 = 64

A = 2

Inradius = A/s = √2/ 3+√5