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Question

Find the radius of the circle which is inscribed in the triangle formed by the straight lines whose equations are 2x+4y+3=0, 4x+3y+3=0, and x+1=0.

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Solution

2x+4y+3=0.....(i)4x+3y+3=0.....(ii)x+1=0........(iii)

Solving (ii) and (iii) we get A(1,13)

solving (iii) and (i) we get B(1,14)

solving (i) and (ii) we get C(310,35)

Distance BC=a

a=(310+1)2+(35+14)2=7520

Distance CA=b

b=(310+1)2+(3513)2=76

Distance AB=c

c=(1+1)2+(13+14)2=712

Coordiantes of incentre are

(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)

⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪(7520×1)+(76×1)+(712×310)7220+76+712⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪,⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪(7520×13)+(76×14)+(712×35)7220+76+712⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎥ ⎥ ⎥ ⎥ ⎥ ⎥

(8575120,21565120)

Radius is equal to distance from x+1=0

=∣ ∣ ∣ ∣8575120+112+02∣ ∣ ∣ ∣
=3575120




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