Question

0(0, 0), A(6, 0) and B(0, 8) are vertices of a triangle. Find the co-ordinates of the incentre of ∆OAB.

Answer 1

Let bisector of ∠O meet AB at point D and bisector of ∠A meet BO at point E 

∴ Point D divides seg AB in the ratio l(OA): l(OB) and point E divides seg BO in the ratio l(AB): l(AO) 

Let I be the incentre of ∠OAB.

∴ Point D divides AB internally in 6 : 8 i.e. 3 :4

∴ -2y = x – 6 

∴ x + 2y = 6 …(ii) 

To find co-ordinates of incentre, we have to solve equations (i) and (ii). 

Substituting y = x in (ii), we get x + 2x = 6

∴ x = 2 Substituting the value of x in (i), we get y = 2 

∴ Co-ordinates of incentre I ≡ (2, 2)

Alternate Method: 

Let I be the incentre. 

I lies in the 1st quadrant. 

OPIR is a square having side length r. 

Since OA = 6, OP = r, 

PA = 6 – r 

Since PA = AQ, 

AQ = 6 – r …(i) 

Since OB = 8, OR = r, 

BR = 8 – r 

∴ BR = BQ 

∴ BQ = 8 – r …(ii)

∴ BQ + AQ= 10 

∴ (8 – r) + (6 – r) = 10 

∴ 2r = 14- 10 = 4 

∴ r = 2 

∴ I = (2,2)