Find Inradius of ΔABC: Coordinates (4,-1) & Internal Bisectors

[utkarshakash]

Homework Statement

In ΔABC the cooridnates of the vertex A are(4,-1) and the lines x-y-1=0 and 2x-y=3 are internal bisectors of angles B and C then radius of incircle of ΔABC is

Homework Equations

The Attempt at a Solution

I can solve the given two eqns to get intersection point which is (2,3). Finding the inradius requires me to calculate the distance from (2,3) to any side. For this I will need eqn of anyone side. Now I can also find the 3rd bisector.

 

[haruspex]

There's probably a better way, but if the radius is r then you can write down the equation for a tangent from A to the circle. This will intercept the given lines at B and C.

 

[utkarshakash]

There's probably a better way, but if the radius is r then you can write down the equation for a tangent from A to the circle. This will intercept the given lines at B and C.

Let us assume that the point of contact of tangent from A is (h,k). Equation of incircle:

$(x-2)^2+(y-3)^2 = r^2$

Equation of tangent from A is $y-k = \dfrac{2-h}{k-3} (x-h)$

Since it passes through A, I get the following condition

$h^2+k^2-2k-6h+5=0$

 

[haruspex]

Posting in the right thread this time:
Try this:
Drop a perpendicular from the incentre O to each of the three sides, meeting AB at C', BC at A' and CA at B'. ∠AOB' = ∠AOC' = α, say; ∠BOA' = ∠BOC' = β, etc. So what does α+β+γ equal? What is ∠BOC in terms of these? You can determine the value of ∠BOC. What does that give for the value of α? Can you use AO and α to find r?