Math, asked by Himanashuyadav, 5 months ago

2: A point equidistant from the lines 4x+3y+10=0 , 5x-12y+26=0 and 7x+24y-50=0 is

A  (1, -1)

B  (1, 1)

C  (0, 0)

D  (0, 1)

Answers

Answered by akhilpradeep2004

1

Answer:

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Step-by-step explanation:

have a great day

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