Question

Find the incenter of triangle formed by the straight line x+1=0,3x-4y=5and 5x+12y=27​

Answer 1

Answer:

By solving the equations of given lines find the coordinates of the vertices as A(3,1),B(−1,−2),C(−1,8/3)

The bisectors of ∠A are given by

3x-4y-5/5=+- 5x+12y-27/13

or x−8y+5=0 and 8x+y−25=0

Since we have to find the internal bisector of angle A, the points B(−1,−2) and C(−1,8/3) must lie on the opposite sides of the internal bisector. Putting in any equation of the bisector say in x−8y+5=0, we get

20,-52/3 i.e of opposit signs and hence this bisector is internal bisector of angle A. It may be verified that B and C when put in 2nd bisector will give results of the same sign. Similarly the internal bisector of angle C is found to be 9x+6y−7=0. Solving these two internal bisectors, we get the coordinates of incentre as (1/3,2/3)

Answer 2

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