The incentre of the triangle formed by the lines 4x+3y+10=0, 5x-12y+26=0, 7x+24y-50=0 is

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(1, -1)

(0, 0)

(1, 1)

(0, 1)

answer is C.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 4 x + 3 y + 10 = 0 - - (1) \\ 5 x - 12 y + 26 = 0 - - (2) \\ 7 x + 24 y - 50 = 0 - - (3) \\ l e t (a, b) b e t h e p o int E q u i d i s \tan c e f r o m g i v e n l i n e s \\ d_{1} = \frac{| 4 a + 3 b + 10 |}{\sqrt{16 + 9}} = \frac{| 4 a + 3 b + 10 |}{5} \\ d_{2} = \frac{| 5 a - 12 b + 26 |}{\sqrt{25 + 144}} = \frac{| 5 a - 12 b + 26 |}{13} \\ d_{3} = \frac{| 7 a + 24 b - 50 |}{\sqrt{49 + 576}} = \frac{| 7 a + 24 b - 50 |}{25} \\ ∴ \frac{| 4 a + 3 b + 10 |}{5} = \frac{| 5 a - 12 b + 26 |}{13} = \frac{| 7 a + 24 b - 50 |}{25} \\ o n l y a = 0, b = 0 s a t i s f y i n g t h e a b o v e E q u a t i o n \\ ∴ I n c e n t r e = (0, 0) \end{array}$